Jean-Christophe Poutineau
email: jean-christophe.poutineau@univ-rennes1.fr.

Karolina Sobczak
email: karolina.sobczak@ue.poznan.pl.

Gauthier Vermandel
email: gauthier@vermandel.fr

The analytics of the New Keynesian 3-equation Model

Keywords: Dynamic IS curve, impulse response analysis, New Keynesian Macroeconomics, New Keynesian Phillips Curve, output gap, Taylor rule. JEL codes: C63, E12, E32, E52

This paper aims at providing a self-contained presentation of the ideas and solution procedure of New Keynesian Macroeconomics models. Using the benchmark "3 equation model", we introduce the reader to an intuitive, static version of the model before incorporating more technical aspects associated with the dynamic nature of the model. We then discuss the relative contribution of supply, demand and policy shocks to the fluctuations of activity, inflation and interest rates, depending on the key underlying parameters of the economy.

Introduction

Keynesian ideas returned to the forefront of academic research in the mid 90's in a new guise to address questions related to unemployment, economic fluctuations and inflation. This followed a twenty year period that witnessed the domination of new classical ideas on questions about both monetary and real macroeconomics. Before contributing to the building of what is now considered as the workhouse of modern macroeconomics (Carlin and [START_REF] Soskice | Macroeconomics: Institutions, Instability, and the Financial System[END_REF]), the New Keynesian School proposed in the 80's a series of models aimed at providing microeconomic foundations to price and/or wage rigidity 1 and at showing that this key feature of the real world can be explained in a setting with optimizing agents with market power. An important breakthrough was about 15 years ago, with the papers of [START_REF] Goodfriend | The New Neoclassical Synthesis and the Role of Monetary Policy[END_REF] and Clarida, Gali and Gertler [1999]. These contributions introduced a framework mixing Real Business Cycle features with nominal rigidities. This setting now forms the basic analytical structure of contemporaneous macroeconomic models as exemplified by [START_REF] Woodford | Interest and Prices: Foundations of a Theory of Monetary Policy[END_REF] or Gali [2008].

Besides new ideas and a new modelling strategy this New Keynesian Synthesis (NKS) has adopted new solution procedures that may appear cumbersome to nonspecialists. Because of their recursive structure NKS models do not admit a closed form solution but should be solved by borrowing procedures developed for the analysis of stochastic discrete time dynamics systems 2 . The aim of this paper is to provide a compact and self-contained presentation of the structure and of the standard solution procedure of the basic NKS framework known as the "three equation model". We particularly separate the main ideas conveyed by this model, using a static version of the reference framework, from the technical aspects of the solution procedure. In the presentation we emphasise the qualitative similarities between the simple graphical analysis of the static model and the Impulse Response Functions (IRFs) of the model following the occurrence of exogenous shocks. We then illustrate the key features of this model in respect of the analysis of business cycles characteristics.

The paper is organized as follows: In the first section we introduce the general structure of a benchmark NKS model that combines (the log linear versions of) a

Philips' curve, an Euler equation and a monetary policy (Taylor) rule3 . In the second section we set a simple static version of the model to obtain closed form solutions for the key macroeconomic variables and to provide the reader with a graphical analysis of the consequences of demand and supply shocks. In the third section we introduce the Blanchard-Kahn solution procedure to get IRFs and dynamic reactions of the model around a stable steady state following exogenous supply demand and policy shocks. This third section is also devoted to a discussion of business cycles characteristics of the model. Section four concludes.

The 3 equation new Keynesian model

The The New Keynesian Philips' Curve (PC) links current inflation (𝜋 ̂𝑡) to expected future inflation (𝐸 𝑡 {𝜋 ̂𝑡+1 }), to the current output gap (𝑦 ̂𝑡) and to an exogenous supply shock that takes the form of a cost push shock (𝜀 𝑡 𝑆 )4 :

𝜋 ̂𝑡 = 𝛽𝐸 𝑡 {𝜋 ̂𝑡+1 } + 𝜅𝑦 ̂𝑡 + 𝜀 𝑡 𝑆 . (1)
As shown in appendix this relationship comes from the aggregation of the supply decision of firms that have market power and can re-optimize their selling price with discontinuities (i.e. nominal rigidities -they cannot modify their selling price at any point in time). , Thus they set the selling price of their product depending on three main criteria. (i) The first criterion is anticipated inflation: as firms cannot re-optimize their price, they take into account future inflation to set their price today. (ii) The second term is the output gap: when firms set their price they take into account the difference between supply and demand so that inflation reflects stresses on the goods market: firms increase their prices during periods of expansion (𝑦 ̂𝑡 > 0) whilst they decrease it during recessions (𝑦 ̂𝑡 < 0). (iii)

Finally, this relation incorporates a cost push term 𝜀 𝑡 𝑆 (such that 𝜀 𝑡 𝑆 > 0 may indicate an increase in raw materials or energy price in the economy). Actually we assume that 𝜀 𝑡 𝑆 is an AR(1) process5 : 𝜀 𝑡 𝑆 = 𝜌 𝑆 𝜀 𝑡-1 𝑆 + 𝜂 𝑡 𝑆 with 𝜂 𝑡 𝑆 ~𝑁(0, 𝜎 𝑆 2 ) and iid. The New Keynesian Phillips' Curve is derived from the Calvo model [1983] which combines staggered price-setting by imperfectly competitive firms. As presented in the appendix, the Calvo approach assumes that in each period, only a fraction  of firms, randomly chosen, can reset their selling prices6 ). Using this assumption, [START_REF] Clarida | The Science of Monetary Policy: A New Keynesian Perspective[END_REF] show that the Phillips' curve then takes a particularly simple form in which inflation depends on the current gap between actual and equilibrium output as in the standard Phillips' curve but on expected future inflation rather than on past inflation.

The dynamic IS curve is a log linearization of the Euler bond equation that describes the intertemporal allocation of consumption of agents in the economy:

𝑦 ̂𝑡 = 𝐸 𝑡 {𝑦 ̂𝑡+1 } - 1 𝜎 (𝑟̂𝑡 -𝐸 𝑡 {𝜋 ̂𝑡+1 }) + 𝜀 𝑡 𝐷 (2)
In this variable 𝜀 𝑡 𝑅 denotes a monetary policy shock that follows an AR(1) process of the form: 𝜀 𝑡 𝑅 = 𝜌 𝑅 𝜀 𝑡-1 𝑅 + 𝜂 𝑡 𝑅 with 𝜂 𝑡 𝑅 ~𝑁(0, 𝜎 𝑅 2 ) and iid. This shock identifies monetary policy decisions which imply deviations from the standard Taylor rule such as unconventional measures or to reshape the inflation expectations in the medium run. This MP schedule aims at replacing the standard LM curve commonly found in the standard AS-AD model. It proposes an up-to-date description of the behaviour of central banks that control a short run nominal interest rate instead of a monetary aggregate [START_REF] Clarida | The Science of Monetary Policy: A New Keynesian Perspective[END_REF]).

This 3-equation model is a stylised shortcut that encompasses supply and demand relations to determine how the three main macroeconomic variables of interest (the output gap, the inflation rate and the nominal interest rate) react to exogenous supply and demand shocks. In this short presentation we ignore more recent developments associated with the introduction of financial frictions that give rise to an acceleration phenomenon (see for example Poutineau and Vermandel [2015a,b]).

The solution for a static version of the model

This second section simplifies the previous system ( 1)-( 3) to convey the main ideas of the NKS model. Following [START_REF] Bofinger | The BMW Model: A New Framework for Teaching Monetary Economics[END_REF] and Poutineau and Vermandel [2015b] we neglect the dynamic aspects of the model and we concentrate on a static version of the framework 8 . This is helpful to obtain the reduced form for the main variable of interest and to understand intuitions regarding the working of the model using tools similar to the IS-LM and AD-AS frameworks. To obtain the static version of the model we firstly assume that the monetary authorities are perfectly credible in the conduct of monetary policy so that the private sector expects that they reach the targeted inflation rate in future, namely that 𝐸 𝑡 {𝜋 ̂𝑡+1 } = 𝜋 0 , where 𝜋 0 is the long-run targeted rate of inflation. Secondly, we assume that the economy is very close to full employment so that the authorities are able to close the output gap in the future, namely that 𝐸 𝑡 {𝑦 ̂𝑡+1 } = 𝑦 0 . Thus the gap between the real interest rate and the natural interest rate disappears. In this case we can express the monetary policy rule in terms of the real interest rate. Imposing these restrictions of the simplified static framework gives:

𝜋 = 𝜋 0 + 𝜅𝑦 + 𝜀 𝑆 , (4)
𝑦 = 𝑦 0 -𝜎𝑟 + 𝜀 𝐷 , (5)
8 Dynamic aspects will be reintroduced in section 3.

𝑟 = 𝜙 𝜋 (𝜋 -𝜋 0 ) + 𝜙 𝑦 𝑦 + 𝜀 𝑅 . (6)
In equilibrium the values of the output gap 𝑦 * , the inflation rate 𝜋 * and the interest rate 𝑟 * solution to the model ( 4)-( 6) are a linear combination of exogenous shocks:

𝑦 * = 𝑦 0 Ω - 𝜎𝜙 𝜋 Ω 𝜀 𝑆 - 𝜎 Ω 𝜀 𝑅 + 1 Ω 𝜀 𝐷 , 𝜋 * -𝜋 0 = 𝜅 Ω 𝑦 0 - 𝜎𝜅 Ω 𝜀 𝑅 + 𝜅 Ω 𝜀 𝐷 + Ω -𝜅𝜎𝜙 𝜋 Ω 𝜀 𝑆 , 𝑟 * = (𝜙 𝜋 𝜅 + 𝜙 𝑦 )𝑦 0 Ω + 𝜙 𝜋 𝜅 + 𝜙 𝑦 Ω 𝜀 𝐷 + 𝜙 𝜋 Ω 𝜀 𝑆 + 1 Ω 𝜀 𝑅 ,
where Ω = (1 + 𝜎(𝜙 𝜋 𝜅 + 𝜙 𝑦 )).

The adjustment of the output gap, the inflation rate and the nominal interest rate following alternative shocks is summarized in

Ω < 0 1 Ω > 0 -𝜎 Ω < 0 Inflation 𝜕𝜋/𝜕𝜀 1 + 𝜎𝜙 𝑦 Ω > 0 𝜅 Ω > 0 -𝜎𝜅 Ω < 0 Interest rate 𝜕𝑟/𝜕𝜀 𝜙 𝜋 Ω >0 𝜙 𝜋 𝜅 + 𝜙 𝑦 Ω > 0 1 Ω > 0

Table 1: reduced form of the static model

To understand more clearly the reaction of the economy to supply and demand shocks we refer the reader to figures 1 and 2. Graphically the model can be represented as consisting of two panels: in the lower panel of each figure, the IS-MP block (equations ( 5) and ( 6)) presented in the (𝑦, 𝑟) space focuses on demand side aspects and can be treated as a New Keynesian representation of the IS-LM framework; in the upper panel the AD-PC block presented in the (𝑦, 𝜋) space determines the global equilibrium of the economy and can be treated as a New

Keynesian representation of the AD-AS framework. The PC curve is shown by equation ( 4) and the AD curve is obtained by combining equations ( 5) and ( 6) and is defined in equation ( 7),

𝑦 = 𝑦 0 1 + 𝜎𝜙 𝑦 - 𝜎𝜙 𝜋 1 + 𝜎𝜙 𝑦 (𝜋 -𝜋 0 ) + 1 1 + 𝜎𝜙 𝑦 𝜀 𝑡 𝐷 (7)
The consequences of the demand shock are presented in Figure 1 both the initial demand shock and the monetary reactionis less than proportional to the demand shock. Furthermore, with the reaction of the central bank the increase of inflation is dampened. Finally the positive demand shock leads to an increase in the output gap, an increase in inflation and a rise in interest rates, as summarized in Table 1.

The consequences of the supply shock are presented in Figure 2. In this example the supply shock is a positive inflation shock (that corresponds to a decrease in the supply of goods). Following this supply shock the Phillips' curve moves upwards to the left in the Finally, the balance between the consequences of the shocks on activity and inflation depends on the slope of the demand curve which, in turn, is affected by the reaction of the central bank to inflation rate and output gap developments. A more conservative central bank (namely a central bank that puts a higher weight on inflation and a lower weight on the output gap) makes the slope of the demand curve of the economy flatter in the upper panel of figures 1 and 2, which translates into lower fluctuations in the interest rate but to a higher variability of the output gap. Conversely if the stance of the central bank reaction is more sensitive to the output gap and less sensitive to inflation the MP and AD curves become steeper and shocks have a lower impact on activity and a higher impact on inflation.

The Fully-Fledged Model

In the dynamic version of the model ( 1)-(3), each period t corresponds to a quarter.

As the fully fledged model does not allow a closed form solution it must be simulated around a stable steady state. The solution procedure, based on the [START_REF] Blanchard | The Solution of Linear Difference Models under Rational Expectations[END_REF] approach9 , requires the choice of numerical values for the parameters of the model in order to compute Impulse Response Functions (IRFs hereafter) and the corresponding variance decomposition of the three variables of interest of the model.

a. The Solution Procedure

The solution procedure introduced by [START_REF] Blanchard | The Solution of Linear Difference Models under Rational Expectations[END_REF] is based on matrix calculus and is aimed at selecting a unique stable dynamic path to describe the reaction of the variables following the occurrence of exogenous shocks. The Blanchard-Kahn condition defines a necessary criterion to get this result through the equality between the number of forward variables and the number of unstable eigenvalues. Practically the problem of the eigenvalues translates into the problem of appropriate values of the structural parameters of the model or their combinations. To be solved the model has first to be written in a state-space representation. For our linear model ( 1

)-(3), defining   1           y , this representation is:     , ) ( 1 ˆ1 1                                      R t S t D t S t t t t t y t t E y E y                     (8)
The Blanchard-Kahn condition states that there are as many eigenvalues of the matrix

          ) ( 1 y T          Z
greater than one in modulus, as there are nonpredetermined variables. Since there are two forward-looking variables in the model ( 1

b. Impulse-response analysis

The mechanisms by which random innovations change into endogenous variables fluctuations may be illustrated by impulse response functions (IRFs). Each IRF isolates the impact of a particular shock throughout the economy. To document the response of activity, inflation and nominal interest we sequentially describe the consequences of a supply, demand and interest rate shock. 10 Galí, Gertler, López-Salido [2001] and [START_REF] Sbordone | Prices and Unit Labor Costs: Testing Models of Pricing Behavior[END_REF] provide estimations based on aggregate data. Galí [2007] points also to some micro evidence.

11 These values were originally proposed by [START_REF] Taylor | A Historical Analysis of Monetary Policy Rules[END_REF] as a good approximation of the monetary policy conducted by the Federal Reserve in years 1986-1999 when the head of the USA central banking system was Alan Greenspan. His monetary policy decisions largely followed standard Taylor rule criteria.

The demand shock: Figure 3 documents the consequences of a 1% positive demand shock. As observed the increase in goods' demand leads to an increase in activity so that the output gap becomes positive. However, as production overshoots its natural value this rise in activity increases the inflation rate. Since both the output gap and inflation rate increase the central bank should react by raising the nominal interest rate. According to the Taylor principle the nominal interest rate increases more than proportionally to inflation developments to affect real exchange rates. This policy however is not sufficient to close the positive output gap immediately or to dampen the inflation rate. The effect of monetary policy should be assessed over time on the output gap (activity goes back to its natural value as time passes) and on the rate of inflation (that converges towards its natural value). The adjustment time path is affected by the parameter value of the Taylor rule. As presented in Figure 3 a higher concern for inflation or output gap reduces the volatility of both activity and inflation. Thus, stricter monetary policy leads to more moderate responses of variables to the demand shock.

The supply shock: In Figure 4, the IRFs describing the consequences of a 1% increase in inflation (i.e. the negative supply shock acts as an increase in the price of raw materials or energy that increases the real marginal cost of production) can be assessed as follows: This shock has a direct impact on inflation that rises and overshoots its targeted value. As a consequence monetary authorities should react according to the Taylor principle by raising the interest rate. Since the increase in the nominal interest rate is higher than the rate of inflation the real rate rises. This, in turn, negatively affects output that decreases under its natural value. However, as time passes, the increase in the interest rate dampens inflation. Finally, the output gap goes back to its steady state value whilst the inflation rate reaches its

c. Business Cycle Statistics

IRF analysis aims at isolating the effect of a particular shock on the dynamics of endogenous variables. However, in real life situations, shocks occur both randomly and jointly to affect the macroeconomic equilibrium. The combined effect of supply and demand shocks over time is captured by historical variance analysis. The aim of this exercise is both to evaluate the relative contribution of each type of shock on the motion of macroeconomic variables over time and to appreciate how a particular design for economic policy may dampen the effect of one particular type of shock. In panel 2 (sticky economy) and panel 3 (quasi flexible economy) we evaluate the sensitivity of the benchmark results to alternative assumptions regarding nominal rigidities. In the sticky economy only 5% of the total number of firms can reset their price each period. Whilst in the quasi flexible situation 99% of the total number of firms reset their prices each quarter. The main consequences can be assessed with regard to the contribution of supply side shocks to inflation and interest rates. Remarkably supply side shocks have no effect on either inflation or interest rates when prices are flexible. In contrast the fluctuations of the output gap are more sensitive to interest rate shocks whilst the effect of demand shocks on activity is almost unaffected.

In panel 4 and 5 we evaluate the sensitivity of the benchmark results to alternative assumption regarding the conduct of monetary policy. A monetary policy which is more aggressive in terms of inflation (panel 4) dampens the effect of demand shocks on activity (and in contrast makes output development more sensitive to supply shocks) and reinforces the impact of demand shocks on inflation (whilst, conversely, it dampens the impact of supply side shocks on this variable). Finally, this policy has almost no noticeable effect on the relative contribution of shocks on interest rate developments. In panel 5 an output oriented monetary policy increases the effect of supply shocks on inflation and interest rate whilst leaving the relative contribution of shocks on activity almost unchanged. The results obtained in these last two panels may serve as simple guideline to determine the nature of monetary policy depending on both its objective and the origin of shocks. If an economy is mainly affected by price mark-up shocks monetary policy should be more closely oriented towards output developments. As this policy is able to dampen the effect of supply shocks on inflation, whilst having no noticeable effect on activity, monetary authorities are able to stabilise prices more easily. In contrast if the economy is affected by demand shocks the authorities have to use arbitrage as a more aggressive policy to fight inflation which dampens the impact of demand shocks on activity whilst it increases the impact of demand shocks on inflation.

Conclusions

In this paper we have described in a concise way the main ideas conveyed by the

  New Keynesian Synthesis (NKS) mixes the methodology of Real Business Cycles (RBC) with nominal and real rigidities to characterise short run macroeconomic developments. More particularly the NKS seeks to explain the macroeconomic short run evolution of an economy subject to real and monetary shocks and to replicate business cycle statistics. The core representation of this synthesis has given rise to what is called the "3-equation model" as the basic NKS setting reduces to a simple system of three equations corresponding to an AS-AD model. First, the AS curve is represented by the New Keynesian Phillips curve that relates inflation to the output gap. Second, the AD component of the model combines a dynamic IS curve (that relates the evolution of the output gap to the interest rate) and a MP (Monetary Policy) schedule (that describes how the nominal interest rate is set by the central bank following fluctuations in the output gap and in the inflation rate. This model is based on agents' micro founded decision rules where consumers maximize their welfare subject to an intertemporal budget constraint and where firms maximize their profit, subject to nominal rigidities, characterising the imperfect adjustment of prices on the goods market. For convenience the micro foundations of this model and the derivation of the log-linear system are presented in appendix. These equations in turn determine three main variables of interest in a closed economy, namely the output gap (𝑦 ̂𝑡) which is the gap between the effective output and potential output, the inflation rate (𝜋 ̂𝑡) and the nominal interest rate (𝑟̂𝑡). Formally, the model is defined as follows:

  . The first panel displays the adjustment of the inflation rate and the output gap. The second panel displays the adjustment of the demand side, accounting for the reaction of the central bank to the shock.

Figure 1 :

 1 Figure 1: Demand shock Figure 2: Supply shock

  shock leads to an increase in the rate of inflation and the central bank reacts by raising the interest rate. Graphically the reaction of the central bank means increasing the interest rate for any value of the output gap so that the MP curve moves left to MP' in the lower panel of Figure2.Once all the adjustments have been implemented the final equilibrium lies at point B which is characterized by a negative output gap (namely activity falls below its natural value), an increase of the inflation rate over its targeted value and at point B' an increase in the interest rate (needed to dampen part of the inflation consequences of the supply shock).

  we know that there should be exactly two eigenvalues outside the unit circle to get one unique stable trajectory of each of the model's variable around the steady state. Given the form of the matrix , T Z the Blanchard-Kahn condition for the model (1)-(3) reduces to the following relation: 𝜅(𝜙 𝜋 -1) + (1 -𝛽)𝜙 𝑦 > 0.

Figure 3 :

 3 Figure 3: Effects of a 1% demand shock Note: Benchmark regime is: 𝜙 𝜋 = 1.5 , 𝜙 𝑦 = 0.5/4, inflation target regime: 𝜙 𝜋 = 1.7, output gap regime: 𝜙 𝑦 = 0.8/4.

  targeted value. As previously for the demand shock the time path of variables is affected by the parameter values of the Taylor rule. A higher concern for output gap (as represented with 'inflation target' IRF) leads to weaker responses of real variables and stronger responses of nominal variables. Inversely a higher concern for inflation leads to stronger responses of real variables and weaker responses of inflation and nominal interest rate.

Figure 4 :

 4 Figure 4: Effects of a 1% supply shock Note: Benchmark regime is: 𝜙 𝜋 = 1.5 , 𝜙 𝑦 = 0.5/4, inflation target regime: 𝜙 𝜋 = 1.7, output gap regime: 𝜙 𝑦 = 0.8/4.

Figure 5 :

 5 Figure 5: Effects of 1% a monetary policy shock Note: Benchmark regime is: 𝜙 𝜋 = 1.5 , 𝜙 𝑦 = 0.5/4, inflation target regime: 𝜙 𝜋 = 1.7, output gap regime: 𝜙 𝑦 = 0.8/4.

3

  equation New Keynesian model and the main elements of the solution procedure required to analyse the dynamics of the model. To introduce the reader to these types of models we have presented a simple static version of the model that gives both direct reduced forms and provides the basis for a simple graphical analysis of the macroeconomic equilibrium. We have then introduced the Blanchard-Kahn solution procedure and report IRFs to describe the dynamic adjustment of the economy over periods. Finally we have used the historical variance analysis to evaluate how a modification of the value of key parameters of the model affect the relative contribution of supply side and demand side shocks. Our aim was not to provide the reader with a comprehensive and up to date catalogue of all the results obtained by this New Keynesian literature but rather to offer a clear and simple presentation of the basic ideas and the required technical tools needed to solve this type of model that have become the conventional workhorse of today's macroeconomics.

  max

Table 1

 1

		Supply Shock 𝜀 𝑆	Demand Shock 𝜀 𝐷	Monetary Shock 𝜀 𝑅
	Output gap 𝜕𝑦/𝜕𝜀	-𝜎𝜙 𝜋

. As shown in the first column a supply shock leads to a decrease in the output gap, (activity decreases below its natural level), and to an increase in the inflation rate and in the interest rate. As shown in the second column a demand shock leads to an increase in the output gap, (activity increases), in the inflation rate and the interest rate. As observed, the reactions of the variables of interest to exogenous shocks are clearly affected by the value of the parameters of the interest rate rule of the authorities (𝝓 𝝅 and 𝝓 𝒚 ).

Table 2 : Calibration of parameters Source: Authors' synthesis

 2 This condition reduces to the choice of appropriate values for the parameters of the model. A sufficiently relevant condition for the previous one is that the monetary authorities should respond more than proportionally to inflation developments (namely, 𝜙 𝜋 > 1 ) according to the Taylor principle. In this case a rise in inflation leads to a more than proportional rise in nominal interest leading to an increase in real interest rates that affects agents' economic decisions and thus the real macroeconomic equilibrium of the model. The choice of parameters is therefore a main feature of the analysis as it must both represent economic features and contribute to the Blanchard-Kahn condition. As presented in Table2, following[START_REF] Galí | Monetary Policy, Inflation, and the Business Cycle: An Introduction to the New Keynesian Framework[END_REF], we use a calibration of the model parameters that is normally selected in the literature. The intra-temporal elasticity between intermediate goods is set at 6 which implies a steady state mark-up of 20 % in the goods' market corresponding to what is observed in main developed economies.The sensitivity of the inflation rate to changes in the marginal cost (𝜅) is equal to 0.13 roughly. The value of the discount factor set at 0.99 implies that the steady state quarterly interest rate 𝑟 equals one and the steady state real return on financial assets of about 4 percent per year. Average price duration amounts to three quarters which is consistent with empirical evidence 10 . The values of coefficients in the interest rate rule (3) are consistent with variations observed in the data on inflation and the interest rate given in the annual rates11 . Because in our model periods are interpreted as quarters the output gap coefficient has to be divided by 4.

  Table 3 shows the variance decomposition of activity, inflation and the nominal interest rate under the benchmark calibration In the first panel of Table 3 (Benchmark calibration), supply side shocks (namely price mark-up shocks) explain most of the output variability leaving

	only a marginal contribution (around 4%) to demand and interest rate shocks. In
	contrast the variability of the inflation rate is mainly explained by demand and
	monetary policy innovations. Finally, around 2/3 of interest rate variability is
	explained by real supply side shocks.

  𝑒 𝛾𝜀 𝑡 𝑆 is the time-varying mark-up, 𝜖 denotes the imperfect substitutability between different goods varieties, 𝜀 𝑡 𝑆 denotes the mark-up shock and 𝛾 is a shift parameter that normalizes the shock to unity in the log-linear form of the model as in[START_REF] Smets | Comparing shocks and frictions in us and euro area business cycles: a Bayesian dsge approach[END_REF]. Since firms are owned by households they discount the expected profits using the same discount factor as households (𝛽 𝜏 𝜆 𝑡+𝜏 𝑐 /𝜆 𝑡 𝑐 ). The first order condition is thus: 𝑅 𝑡 is the nominal interest rate, 𝜋 𝑡 is the inflation rate, 𝑌 𝑡 is the level of output and 𝜀 𝑡 𝑅 is an AR(1) monetary policy shock. Finally, parameters 𝑅 ̅ , 𝜋 ̅ and

		𝑃 𝑡 * (𝑖)	𝐸 𝑡 ∑ +∞ 𝜏=0	𝑐 𝜆 𝑡+𝜏 𝜆 𝑡 𝑐 (𝛽𝜃) 𝜏	[𝑃 𝑡 * (𝑖) -𝑀𝐶 𝑡+𝜏 (𝑖)]𝑌 𝑡+𝜏 (𝑖),	(A.15)
	under the downward sloping constraint from goods' packers:
			𝑌 𝑡+𝜏 (𝑖) ( 𝑃 𝑡 * (𝑖) 𝑃 𝑡+𝜏 ) -𝜇 𝑡+𝜏 -1 𝜇 𝑡+𝜏	𝑌 𝑡+𝜏 ,	𝜏 > 0,	(A.16)
	where 𝜇 𝑡 =	𝜖 𝜖-1
	𝐸 𝑡 ∑ +∞ 𝜏=0	𝜆 𝑡+𝜏 𝑐 𝜆 𝑡 𝑐	(𝛽𝜃) 𝜏 𝜇 𝑡+𝜏 -1	[𝑃 𝑡 * (𝑖) -𝜇 𝑡+𝜏 𝑀𝐶 𝑡+𝜏 (𝑖)]𝑌 𝑡+𝜏 (𝑖) = 0.	(A.17)
	A.3. Authorities
	To close the model the monetary policy authority sets its interest rate according
	to a standard Taylor Rule:
				𝑅 𝑡 𝑅 ̅	= (	𝑅 𝑡 𝑅 ̅	)	𝜌	((	𝜋 𝑡 𝜋 ̅	) 𝜙 𝜋	(	𝑌 𝑡 𝑌 ̅	)	𝜙 𝑦	1-𝜌 )	𝑒 𝜀 𝑡 𝑅 ,	(A.18)
	where

In the appendix, we provide the micro foundations of the framework used in this paper.

In the paper all parameters are positive.

This assumption is generally adopted in the literature to characterize exogenous shocks (see for example, Gali[2008] for a discussion)

[START_REF] Baranowski | Inflacja a mechanizmy aktualizacji cen. Studium dla Polski[END_REF] propose an endogenous mechanism.

In appendices, we provide an interest rate smoothing, in this section, we neglect these features such that 𝜌 = 0.

In this paper we adopted the Blanchard-Kahn approach for solving the model, given its anteriority and popularity in literature. However, the reader should be aware of the existence of other methods introduced by[START_REF] Klein | Using the generalized Schur form to solve a multivariate linear rational expectations model[END_REF] and[START_REF] Sims | Solving linear rational expectation models[END_REF].[START_REF] Miao | Economic Dynamics in Discrete Time[END_REF] offers a nice comparison between these three approaches.

of table 2 and evaluates the sensitivity of the benchmark results to alternative values of key behavioural and policy parameters of the model.

where 𝜎 > 0 and 𝜑 > 0 are shape parameters of the utility function with respect to consumption and to labour supply whilst 𝜒 is the a shift parameter which scales the steady state labour supply to realistic values. As in [START_REF] Smets | Comparing shocks and frictions in us and euro area business cycles: a Bayesian dsge approach[END_REF] we introduce an AR(1) demand shock process in the budget constraint of the representative household denoted 𝜀 𝑡 𝐷 .

After replacing the Lagrange multiplier the first order conditions are defined by the Euler bond condition:

where 𝜋 𝑡+1 = 𝑃 𝑡+1 /𝑃 𝑡 is the inflation rate and the labour supply equation is determined by: 𝜒𝐶 𝑡 (𝑗) 𝜎 𝐻 𝑡 (𝑗) 𝜑 = 𝑊 𝑡 𝑃 𝑡 (A.11)

These equations define the optimal paths of labour and consumption and maximize the welfare index of the representative household.

A.2. Firms

The representative firm 𝑖 maximizes its profits:

𝑃 𝑡 (𝑖)𝑌 𝑡 (𝑖) -𝑊 𝑡 𝐻 𝑡 (𝑖), (A.12)

under the supply constraint:

We suppose that firms solve a two-stage problem. In the first stage, firms choose labour demand in a perfectly competitive market. The first order condition is:

where 𝑀𝐶 𝑡 denotes the nominal marginal cost of producing one unit of goods.

In the second stage problem the firms cannot optimally set prices. There is a fraction of firms 𝜃 that are not allowed to reset prices , prices then evolve according to 𝑃 𝑡 (𝑖) = 𝑃 𝑡-1 (𝑖) whilst for the remaining share of firms 1 -𝜃, they are able to set their selling price such that 𝑃 𝑡 (𝑖) = 𝑃 𝑡 * (𝑖), where 𝑃 𝑡 * (𝑖) denotes the optimal price set by the representative firm given the nominal rigidity. The maximization programme is thus defined as:

𝑌 ̅ are steady state values for the interest rate, the inflation rate and GDP 12 . The central bank reacts to the deviation of the inflation rate and the GDP from their steady state values in a proportion of 𝜙 𝜋 and 𝜙 𝑦 , the central bank also smoothes its rate at one degree 𝜌.

A.4. Equilibrium conditions

After aggregating all the varieties supplied by firms the resource constraint for the economy is defined by:

Whilst the aggregation between constrained firms and non-constrained firms leads to the following equation for aggregate prices:

B -Linearization

To obtain the steady state of the model, we normalize prices 𝑃 = 1 whilst we assume that households work one third of their time 𝐻 = 1/3. Then we find:

First, combining the Euler bond equation (A.10) and the resources constraint (A.19), i.e. 𝑦 ̂𝑡 = 𝑐̂𝑡, then production is determined by:

12 Under a credible central bank, 𝜋 ̅ and 𝑌 ̅ also can be interpreted as the targets of the central bank in terms of inflation rate and GDP.

𝑦 ̂𝑡 = 𝑦 ̂𝑡+1 -1 𝜎 (𝑟̂𝑡 -𝐸 𝑡 𝜋 ̂𝑡+1 ) + 𝜀 𝑡 𝐷 .

(A.21)

The labour supply equation (A.11) in log-deviation is: (A.22) where 𝑤 ̂𝑡 denotes the variations of the real wage. Up to a first order approximation of the firm price optimization solution (A.17


Nicolas Chevallier

Costin Augustin Fruchard

C Vîlcu

Envelopes of α-sections

Keywords: Convex body, alpha-section, envelope, floating body, fair partitioning. MSC Classification: 52A10, 52A38, 51M25, 51M04

published or not. The documents may come

Introduction

In this paper, unless explicitly stated otherwise, K denotes a convex body in the Euclidean plane E; i.e., a compact convex subset of E with nonempty interior. Let ∂K denote the boundary of K and |K| its area. Given α ∈ ]0, 1[, an α-section of K is an oriented line ∆ ⊂ E cutting K in two parts, one to the right, denoted by K -, of area |K -| = α|K|, and the other to the left, K + , of area |K + | = (1α)|K|; here K ± are compact sets, thus

K + ∩ K -= ∆ ∩ K.
Denote by K α the intersection of all K + and call it the α-core of K; denote by m α the envelope of all α-sections of K.

The purpose of this article is to study m α and its relation to the α-core. We refer to Sections 6-7 for formal statements, and give in this introductory section an informal presentation. Since K α is empty for α > 1 2 , we will often implicitly assume α ≤ 1 2 when dealing with α-cores. The situation depends essentially upon whether K is centrally symmetric (we will say 'symmetric' in the sequel, for short) or not. Precisely, we prove the following statements.

If K is symmetric then one has m α = ∂K α for all α ∈ 0, 1 2 . Moreover, we have the following equivalence: the envelope m α is of class C 1 for all α ∈ 0, 1 2 if and only if K is strictly convex.

If K is non-symmetric then we cannot have m α = ∂K α for all α ∈ 0, 1 2 , because m α exists for all α, whereas K α is empty for α close enough to 1 2 . More precisely, there exists a critical value α B ∈ 0, 1 2 such that for all α ∈ ]0, α B ] we have m α = ∂K α , and for all α ∈ α B , 1 2 we have m α ∂K α . The case α B = 0 can occur, e.g., if there exists a triangle containing K with an edge entirely contained in ∂K. We also prove that m α is never of class C 1 for α ∈ α B , 1 2 , and that m α is of class C 1 for all α ∈ ]0, α B [ if and only if ∂K does not contain two parallel segments. As a by-product, we obtain the following characterization: A convex body K is non-symmetric if and only if there exists a triangle containing more than half of K (in area), with one side entirely in K and the two others disjoint from the interior of K.

conjecture is as follows: If K ⊆ L are two convex bodies and α ∈ 0, 1 2 , then there exists an α-section of L which either does not cross the interior of K, or is a β-section of K for some β ≤ α. The conjecture has been recently proven in the case of planar convex bodies, see [START_REF] Fruchard | Fair partitioning by straight lines[END_REF].

There are two reasons which motivated us to undertake this systematic study of αsections and their envelopes. First, we found only one reference which focuses especially on the envelope of α-sections, Lecture 11 of the nice book [START_REF] Fuchs | Mathematical omnibus: Thirty lectures on classic mathematics[END_REF], which is a simplified approach. Although the results contained in the present article seem natural, and the proofs use only elementary tools and are most of the time simple, we hope that our work will be helpful in clarifying the things.

Our second motivation for studying α-sections and their envelope was a problem of fair partitioning of a pizza. What we call a pizza is a pair of planar convex bodies K ⊆ L, where L represents the dough and K the topping of the pizza. The problem of fair partitioning of convex bodies in n pieces is a widely studied topic, see e.g. [START_REF] Bárány | Equipartitioning by a convex 3-fan[END_REF][START_REF] Bárány | Simultaneous partitions of measures by k-fans[END_REF][START_REF] Bereg | Orthogonal equipartitions[END_REF][START_REF] Blagojević | Convex equipartitions via equivariant obstruction theory[END_REF][START_REF] Karasev | Equipartition of several measures[END_REF][START_REF] Karasev | Convex equipartitions: the spicy chicken theorem[END_REF][START_REF] Sakai | Balanced convex partitions of measures in R 2[END_REF][START_REF] Soberón | Balanced convex partitions of measures in R d[END_REF]. Nevertheless, to our knowledge, our way of cutting has never been considered: We use a succession of double operations: a cut by a full straight line, followed by a Euclidean move of one of the resulting pieces; then we repeat the procedure. The final partition is said to be fair if each resulting slice has the same amount of K and the same amount of L. The result of [START_REF] Fruchard | Fair partitioning by straight lines[END_REF] is the following: Given an integer n ≥ 2, there exists a fair partition of any pizza (K, L) into n parts if and only if n is even.

Notation, conventions, and preliminaries

The notation S 1 stands for the standard unit circle, S 1 := R/(2πZ), endowed with its usual metric d(θ, θ ′ ) = min{|ττ ′ | ; τ ∈ θ, τ ′ ∈ θ ′ }. On S 1 we use the notation: θ ≤ θ ′ if there exist τ ∈ θ, τ ′ ∈ θ ′ such that τ ≤ τ ′ < τ + π; it is not an order because it is not transitive.

Given θ ∈ S 1 , let u(θ) denote the unit vector of direction θ, u(θ) = (cos θ, sin θ). For convenience, we add arrows on vectors. Unless explicitly specified otherwise, all derivatives will be with respect to θ, hence e.g., u ′ (θ) = d u dθ (θ) = (-sin θ, cos θ) is the unit vector orthogonal to u(θ) such that the frame u(θ), u ′ (θ) is counterclockwise.

Given an oriented straight line ∆ in the plane, ∆ + denotes the closed half-plane on the left bounded by ∆, and ∆ -is the closed half-plane on the right. We identify oriented straight lines with points of the cylinder C = S 1 × R, associating each pair (θ, t) ∈ C to the line oriented by u(θ) and passing at the signed distance t from the origin. In other words, the half-plane ∆ + is given by ∆

+ = {x ∈ R 2 ; x, u ′ (θ) ≥ t}. We endow C with the natural distance d (θ, t), (θ ′ , t ′ ) = d(θ, θ ′ ) 2 + |t -t ′ | 2 ) 1/2 . Given α ∈ ]0, 1[, an α-section of K is an oriented line ∆ such that |∆ -∩ K| = α|K|.
For all α ∈ ]0, 1[ and all θ ∈ [0, 2π[, there exists a unique α-section of K of direction θ; it will be denoted by ∆(α, θ). This defines a continuous function ∆ : ]0, 1[ ×S 1 → C. We obviously have the symmetry

∆ ± (1 -α, θ) = ∆ ∓ (α, θ + π). (1)
For 0 < α ≤ 1 2 , we call α-core of K, and denote by K α , the intersection of all left half-planes bounded by α-sections:

K α = θ∈S 1 ∆ + (α, θ).
It is a compact convex subset of the plane, possibly reduced to one point or empty.

Let ∆ = ∆(α, θ) be an α-section of K of direction u = u(θ), and let b, c denote the endpoints of the chord ∆ ∩ K, with bc having the orientation of u (i.e., the scalar product bc, u is positive), see Figure 1.

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 X 2 2 X u h h h y u ′ q r b ′ b β r m £ £ £ £ # r c ′ c γ K ∂K ∆ ∆ - ∆ + Figure 1: Some notation Let m = 1 2 (b + c
) denote the midpoint of the chord bc. Let h denote the half-length of bc; hence we have b = mh u and c = m + h u. The functions b, c, m and h are continuous with respect to α and θ. As we will see in the proof of Proposition 4.1, they are also leftand right-differentiable with respect to θ at each θ 0 ∈ S 1 , e.g., the following limit exists (with the convention θ → θ -

0 for θ → θ 0 , θ < θ 0 ) b ′ l (θ 0 ) = lim θ→θ - 0 1 θ-θ 0 b(α, θ) -b(α, θ 0 ) ,
and similarly for b ′ r , c l ′ , c r ′ . They are also left-and right-differentiable with respect to α, but we will not use this fact.

We say that b is a regular point of ∂K (regular for short) if there is a unique supporting line to K at b (i.

= ( u, c ′ ) ∈ ]0, π[ (if c is regular), resp. γ l , γ r (if c
is a corner), be the angle between ∆ and the tangent, resp. left-tangent, right-tangent, to ∂K at c, see Figure 1.

For a fixed α, the values of θ such that b(α, θ) or c(α, θ) (or both) is a corner will be called singular; those for which both b and c are regular will be called regular.

Observe that we always have β r ≤ β l and γ l ≤ γ r , with equality if and only if b, resp. c, is regular. Also observe the following fact.

b and c admit parallel supporting lines if and only if

β r + γ l ≤ π ≤ β l + γ r . (2)
Finally, observe that the angle between b ′ , (resp. b ′ l , b ′ r ) and the axis of abscissae is equal to θβ(α, θ) (resp. θβ l (α, θ), θβ r (α, θ)). Since these angles are increasing and intertwining functions of θ, we have the following statement.

If θ < θ ′ then θ -β l (α, θ) ≤ θ -β r (α, θ) ≤ θ ′ -β l (α, θ ′ ) ≤ θ ′ -β r (α, θ ′ ). It follows that lim θ→θ + 0 β l (α, θ) = lim θ→θ + 0 β r (α, θ) = β r (α, θ 0 ) ≤ β l (α, θ 0 ) = lim θ→θ - 0 β l (α, θ) = lim θ→θ - 0 β r (α, θ), (3)
and similarly lim

θ→θ - 0 γ l (α, θ) = lim θ→θ - 0 γ r (α, θ) = γ l (α, θ 0 ) ≤ γ r (α, θ 0 ) = lim θ→θ + 0 γ l (α, θ) = lim θ→θ + 0 γ r (α, θ). (4)
In a same way we have

If α < α ′ then β r (α, θ) ≤ β l (α, θ) ≤ β r (α ′ , θ) ≤ β l (α ′ , θ) and γ l (α, θ) ≤ γ r (α, θ) ≤ γ l (α ′ , θ) ≤ γ r (α ′ , θ). (5)
As a consequence, lim

α→α - 0 β l (α, θ) = lim α→α - 0 β r (α, θ) = β r (α 0 , θ) ≤ β l (α 0 , θ) = lim α→α + 0 β l (α, θ) = lim α→α + 0 β r (α, θ), lim α→α - 0 γ l (α, θ) = lim α→α - 0 γ r (α, θ) = γ l (α 0 , θ) ≤ γ r (α 0 , θ) = lim α→α + 0 γ l (α, θ) = lim α→α + 0 γ r (α, θ).
All these elements b, c, m, h, β, γ can be considered as functions of both α and θ. Nevertheless, as already said, all derivatives are with respect to θ.

Let v = v(α, θ) be the scalar product v = m ′ , u ∈ R.
If v has a discontinuity, then v l and v r denote its corresponding left and right limits. As we will see, m ′ is always collinear to u, hence v is the "signed norm" of m ′ . We will also see that v has discontinuities only if b or c (or both) is a corner. Since we chose θ as parameter, we will also see that v is the signed radius of curvature of the curve m, but we prefer to refer to it as the velocity of the current point m of the envelope. The symmetry [START_REF] Bárány | Equipartitioning by a convex 3-fan[END_REF]

gives m(α, θ + π) = m(1 -α, θ), hence m ′ (α, θ + π) = m ′ (1 -α, θ). Since u(θ + π) = -u(θ), we obtain v l (α, θ + π) = -v l (1 -α, θ) and v r (α, θ + π) = -v r (1 -α, θ). (6)
Our last notation is V for the segment of endpoints v l and v r ; i.e.,

V = [v l , v r ] if v l ≤ v r , V = [v r , v l ] otherwise. Formulae (6) yield V (α, θ + π) = -V (1 -α, θ). ( 7
)
3 A "digressing tour"

Before carrying on α-sections, we would like to digress for a moment in a more general framework. The notion and results of this section are elementary and probably already known, but we did not find any reference in the literature. They can be considered as exercises in a graduate course on planar curves. Recall that a ruled function (or regulated function) R : S 1 → R is the uniform limit of piecewise constant functions. It is equivalent to saying that R admits a left-and a right-limit, denoted below by R l and R r , at each point of S 1 , see e.g., [START_REF] Dieudonné | Foundations of Modern Analysis[END_REF]. Recall also our notation u(θ) = (cos θ, sin θ). Definition 3.1 . (a) A tour is a planar curve parametrized by its tangent. More precisely, we call m : S 1 → R 2 a tour if m is continuous, has a left-and a right-derivative at each point of S 1 , and if there exists a ruled function R :

S 1 → R such that m ′ l (θ) = R l (θ) u(θ) and m ′ r (θ) = R r (θ) u(θ). (b) The core K = K(m)
of a tour m is the intersection of all left half-planes delimited by all tangents to m oriented by u, i.e., K = θ∈S 1 D + (θ), where D(θ) = m(θ) + R u(θ).

Tours are not necessarily simple curves. The case m(θ) = m(θ + π) gives rise to a double half-tour. This is the case e.g. for the envelope of half-sections of a planar convex body. (a) For each θ ∈ S 1 , R l (θ), resp. R r (θ), is the signed left-, resp. right-radius of curvature of m * at m(θ).

(b) If R l and R r do not vanish on S 1 , then m * is a C 1 submanifold of the plane.

(c) Conversely, if there exist θ 1 ≤ θ 2 ≤ θ 1 + π ∈ S 1 such that the product R r (θ 1 )R l (θ 2 ) is negative, then there exists θ 3 ∈ [θ 1 , θ 2 ] such that m * is not of class C 1 at m(θ 3 ). (d)
The core K of m is either empty, or a point, or a strictly convex body.

(e) The boundary of the core, ∂K, is included in m * . Moreover, ∂K and m * coincide if and only if the functions R l and R r are nonnegative on S 1 .

Remarks.

1. In the context of convex floating bodies, the statement (d) above is already known, also in arbirtrary dimension, see e.g. Prop. 1 (iv) and (v) in [START_REF] Werner | Floating bodies and illumination bodies[END_REF] or Theorem 3 in [START_REF] Meyer | A geometric property of the boundary of symmetric convex bodies and convexity of flotation surfaces[END_REF].

2. In the case where R vanishes without changing of sign, m * may, or may not, be of class C 1 .

Proof . (a) Let s denote the curvilinear abscissa on m from some starting point, say m(0).

We have s(θ) = θ 0 R l (τ )dτ = θ 0 R r (τ )dτ . Locally, if R l (θ 0 ) and R r (θ 0 ) are nonzero,
then the tangent of m at m(θ 0 ) is u(θ 0 ) and we have (c) We may assume, without loss of generality, that R r (θ 1 ) > 0 and R l (θ 2 ) < 0. If θ 1 = θ 2 then m(θ 1 ) is a cusp (i.e., a point with one half-tangent), hence m * is not C 1 at m(θ 1 ). In the sequel we assume

d u ds l/r (θ 0 ) = d u dθ dθ ds l/r (θ 0 ) = 1 R l/r (θ 0 ) u ′ (θ 0 ).
θ 1 < θ 2 . Consider E = {θ ∈ [θ 1 , θ 2 ] ; R r (θ) > 0}. Let θ 3 = sup E. Since 0 > R l (θ 2 ) = lim θ→θ - 2 R l (θ) = lim θ→θ - 2 R r (θ), we have θ / ∈ E for θ < θ 2 , θ close enough to θ 2 .
In the same manner, we have θ ∈ E if θ > θ 1 , θ close enough to θ 1 . Therefore, we have θ 1 < θ 3 < θ 2 . Now two cases may occur. If there exists θ 4 > θ 3 , such that R vanishes on the whole interval ]θ 3 , θ 4 [ , then m is constant on [θ 3 , θ 4 ]. We assume θ 4 ≤ θ 2 maximal with this property. By contradiction, if m * is C 1 at the point m(θ 3 ) = m(θ 4 ) then necessarily we have θ

4 = θ 3 + π, a contradiction with θ 1 < θ 3 ≤ θ 4 ≤ θ 1 + π.
In the other case, for all δ > 0 there exists θ ∈ ]θ 3 , θ 3 + δ[ such that R r (θ) < 0, and we obtain that m(θ 3 ) is a cusp.

Before the proofs of (d) and (e), we first establish two lemmas.

Int (K) = θ∈S 1 Int D + (θ) . (8)
Furthermore, we have ∂K ⊂ θ∈S 1 D(θ).

Proof . The inclusion ⊆ in ( 8) is evident: For each θ we have

K ⊂ D + (θ), hence Int (K) ⊂ Int D + (θ) . Conversely, given x ∈ K, the map θ → dist x, D(θ) is continuous. Since S 1 is compact, if x is in the interior of D + (θ) for all θ ∈ S 1
, then this map has a minimum ρ > 0, and the disc of center x and radius ρ is included in K. This proves [START_REF] Dieudonné | Foundations of Modern Analysis[END_REF]. If x ∈ ∂K, then x is in every closed half-plane D + (θ) but not in every open one by ( 8), hence x has to be on (at least) one of the lines D(θ).

Lemma 3.4 . If θ 1 ∈ S 1 and z are such that z ∈ ∂K ∩ D(θ 1 ) then z = m(θ 1 ).
Proof . For small ε = 0, positive or negative, we have

m(θ 1 + ε) -m(θ 1 ) = θ 1 +ε θ 1 R(θ) u(θ)dθ.
By boundedness of R l and R r , and by continuity of u, we deduce that there exists r equal to R l (θ 1 ) or R r (θ 1 ) such that

m(θ 1 + ε) -m(θ 1 ) = r u(θ 1 )ε + o(ε). (9)
It follows that D(θ 1 ) and D(θ 1 +ε) cross at a distance O(ε) from m(θ 1 ), see Figure 3 below. If z were different from m(θ 1 ) then, for ε small, either negative or positive depending on the relative positions of z and m(θ 1 ), we would have

z ∈ Int D -(θ 1 + ε) , hence z / ∈ K, a contradiction.
Now we return to the proof of Proposition 3.2.

(d) By contradiction, assume that ∂K contains some segment [x, y] with x = y and take z ∈ ]x, y[ arbitrarily. By Lemma 3.3, there exists θ 1 ∈ S 1 such that z ∈ D(θ 1 ). Since both x and y belong to K ⊂ D + (θ 1 ), the line D(θ 1 ) contains both x and y, i.e., θ 1 is the direction of ± → xy. By Lemma 3.4, we deduce that z = m(θ 1 ). Hence we proved that any z ∈ ]x, y[ coincides with z = m(θ 1 ), where θ 1 is one of the directions ± → xy, which is impossible.

(e) The first assertion follows directly from Lemmas 3.3 and 3.4. For the second one, first we proceed by contradiction and we assume R l (θ 1 ) < 0 or R r (θ 1 ) < 0 for some θ 1 ∈ S 1 , say R r (θ 1 ) < 0. Then by [START_REF] Dupin | Applications de Géometrie et de Mécanique[END_REF], for ε > 0 small enough, we have m(θ

1 + ε) = E r r I r m(θ 1 ) m(θ 1 + ε) D(θ 1 + ε) D(θ 1 ) z
m(θ 1 ) + εR r (θ 1 ) u(θ 1 ) + o(ε), hence m(θ 1 ) ∈ Int D -(θ 1 + ε) , see Figure 3. It follows that m(θ 1 ) / ∈ K. Conversely, if R l and R r are nonnegative, take θ 0 ∈ S 1 arbitrarily. Then, for all θ ∈ [θ 0 , θ 0 + π], we have m(θ) = m(θ 0 ) + θ θ 0 m ′ (τ )dτ = m(θ 0 ) + θ θ 0 R l (τ ) u(τ )dτ. Therefore we obtain → m(θ 0 )m(θ), u ′ (θ 0 ) = θ θ 0 R l (τ ) u(τ ), u ′ (θ 0 ) dτ ≥ 0, since u(τ ), u ′ (θ 0 ) = sin(θ -θ 0 ) ≥ 0. In the same manner, we have for all θ ∈ [θ 0 -π, θ 0 ], → m(θ 0 )m(θ), u ′ (θ 0 ) ≥ 0, hence the whole curve m * is in the half-plane D + (θ 0 ). This holds for all θ 0 ∈ S 1 , hence m * ⊂ K. Finally, for each θ ∈ S 1 , since m(θ) ∈ D(θ) and K ⊂ D + (θ), m(θ) cannot be in Int (K), hence m(θ) ∈ ∂K.

Dependence with respect to θ

We begin this section with some known results; we give the proofs for the sake of completeness of the article. See Section 2, especially Figure 1, for the notation. We recall that m is the midpoint of the chord bc and that v l|r = m ′ l|r , u .

m ′ = v u and v = h 2 (cotanβ + cotanγ). ( 10
)
In the case where b or c (or both) is a corner of ∂K, we have Statement (a) is attributed to M. M. Day [START_REF] Day | Polygons circumscribed about closed convex curves[END_REF] by S. Tabachnikov. Formula (10) appears in a similar form in [START_REF] Gutkin | Caustics for inner and outer billiards[END_REF].

m ′ l = v l u, m ′ r = v r u, v l = h 2 (cotanβ l + cotanγ l ), and v r = h 2 (cotanβ r + cotanγ r ). (c)
Proof . We fix α ∈ ]0, 1[ and θ ∈ S 1 ; we will not always indicate the dependence in α and θ of the functions b, c, m, u, etc. Let ε > 0, set M (ε) = ∆(α, θ) ∩ ∆(α, θ + ε) (see Figure 4), and consider the curvilinear triangles We have

T b (ε) = ∆(α, θ) -∩ ∆(α, θ + ε) + ∩ K , T c (ε) = ∆(α, θ) + ∩ ∆(α, θ + ε) -∩ K.
|∆(α, θ) -∩ K| = |∆(α, θ + ε) -∩ K| = α, hence 0 = |T b (ε)| -|T c (ε)| = 1 2 ε b -M (ε) 2 -c -M (ε) 2 + o(ε).
As a consequence, we obtain lim

ε→0 + M (ε) = m. The case ε < 0 is similar. Now we prove b ′ r = h cotanβ r u -h u ′ . ( 11
)
For ε > 0, let B(ε) denote the intersection of the right-tangent to ∂K at b with the line ∆(α, θ + ε), see Figure 4. By definition of right-tangent, we have

B(ε) = b(θ + ε) + o(ε). Set H(ε) = b -M (ε) , H 1 (ε) = B(ε) -b, u , H 2 (ε) = M (ε) -B(ε), u , and H 3 (ε) = b -B(ε), u ′ . We obtain the following linear system in H 1 , H 2 , H 3 H(ε) = H 1 (ε) + H 2 (ε), H 3 (ε) = H 1 (ε) tan β r = H 2 (ε) tan ε, giving B(ε) = b - tan β r tan ε tan β r + tan ε H(ε) u + tan ε tan β r + tan ε H(ε) u ′ .
As a consequence, we obtain

B(ε) = b -εH(ε) u + εcotanβ r H(ε) u ′ + o(ε). Since H(ε) = h+o(1), it follows that b(θ+ε) = B(ε)+o(ε) = b-εh u+εcotanβ r h u ′ +o(ε), yielding (11). Similarly, we have b ′ l = h cotanβ l u-h u ′ , c ′ l = h cotanγ l u+h u ′ , and c ′ r = h cotanγ r u+h u ′ . Statements (a) and (b) now follow from m = 1 2 (b + c).
Since the expression of v given by ( 10) is a ruled function, statement (c) follows directly from Proposition 3.2 (a). Corollary 4.2 . For all θ 0 ∈ S 1 and all α ∈ ]0, 1[ , we have

lim θ→θ - 0 v l (θ) = lim θ→θ - 0 v r (θ) = v l (θ 0 ) and lim θ→θ + 0 v l (θ) = lim θ→θ + 0 v r (θ) = v r (θ 0 ). ( 12
)
In other words, in the sense of the Pompeiu-Hausdorff distance, we have

lim θ→θ - 0 V (θ) = {v l (θ 0 )} and lim θ→θ + 0 V (θ) = {v r (θ 0 )}. ( 13
)
Therefore, the function V is upper semi-continuous (for the inclusion) with respect to θ.

Proof . Immediate, using (3), ( 4), [START_REF] Fruchard | Fair partitioning by straight lines[END_REF], and the continuity of the cotangent function.

The next statement will be used both in Sections 5 and 6.

Corollary 4.3 . For θ 1 , θ 2 ∈ S 1 , with θ 1 ≤ θ 2 ≤ θ 1 + π, we have conv V (α, θ 1 ) ∪ V (α, θ 2 ) ⊆ θ 1 ≤θ≤θ 2 V (α, θ). (14)
Proof . Let v be in the above convex hull. If v ∈ V (α, θ 1 ) ∪ V (α, θ 2 ), there is nothing to prove. Otherwise assume, without loss of generality, that max V (α, θ 1 ) < min V (α, θ 2 ) and set θ 0 = sup{θ ≥ θ 1 ; max V (α, θ) < v}. We have θ 1 ≤ θ 0 < θ 2 ; for all θ ∈ ]θ 0 , θ 2 [, max V (α, θ) ≥ v; and there exists a convergent sequence {θ n } n∈N tending to θ 0 with max V (α, θ n ) < v. By [START_REF] Grünbaum | Continuous families of curves[END_REF], we obtain max

V (α, θ 0 ) = v, hence v ∈ V (α, θ 0 ).
5 The forwards, the backwards, and the zero sets

In this section we fix α ∈ ]0, 1[ . Recall that θ is called regular if ∂K is C 1 at b(α, θ
) and c(α, θ), and singular otherwise.

Definition 5.1 . (a) Let F (α) be the set of all θ ∈ S 1 such that either v(α, θ) > 0 (if θ is regular), or V (α, θ) ∩ ]0, +∞[ = ∅ (if θ is singular); we call it the forwards set. (b) Similarly, let B(α) be the set of all θ ∈ S 1 such that either v(α, θ) < 0 (if θ is regular), or V (α, θ) ∩ ] -∞, 0[ = ∅ (if θ is singular)
; we call it the backwards set.

(c) Finally, let Z(α) denote the set of all θ ∈ S 1 such that either v(α, θ) = 0 (if θ is regular), or V (α, θ) contains 0 (if θ is singular); we call it the zero set.

By the symmetry (7), we have (with the notation [θ

1 , θ 2 ] = {θ ∈ S 1 ; θ 1 ≤ θ ≤ θ 2 } for θ 1 , θ 2 ∈ S 1 ) θ ∈ Z(α) ⇔ θ + π ∈ Z(1 -α) and θ ∈ B(α) ⇔ θ + π ∈ F (1 -α). (15)
By Proposition 4.1 (b), we have

θ ∈ Z(α) ⇔ π ∈ [β l + γ l , β r + γ r ] (or π ∈ [β r + γ r , β l + γ l ]).
Since β l ≥ β r and γ r ≤ γ l , by (2), we deduce that, if θ ∈ Z(α) then b = b(α, θ) and c = c(α, θ) admit parallel supporting lines of K. If one of the points b, c or both is regular, the converse is also true; however, it can occur that β r + γ l ≤ π ≤ β l + γ r but π does not belong to [β l + γ l , β r + γ r ] (or to [β r + γ r , β l + γ l ]), and then V = [v l , v r ] (or [v r , v l ]) does not contain 0.

In the case where ∂K is C 1 , by Proposition 4.1, θ belongs to B(α) if and only if β(α, θ) + γ(α, θ) > π, i.e., there exists a triangle T = abc, with one edge equal to the chord bc (with b = b(α, θ), c = c(α, θ)), the other two edges, ab and ac, not crossing the interior of K, and which contains an amount 1α of K: |T ∩ K| = (1α)|K|. In the case where ∂K is not C 1 , this latter condition is necessary but not always sufficient to have θ ∈ B(α). However, we will see in Section 6 that this condition implies θ ∈ B(α ′ ) for all α ′ > α.

Let us observe Z(α) and B(α), denoted Z and B here. These sets can be very complicated, even if ∂K is C 1 . In Section 8 we present a construction which, to any prescribed closed subset C of S 1 , associates a C 1 convex body K such that Z 1 2 coincides with C, up to countably many isolated points. We give next a brief description in the simplest cases.

If θ 0 is an isolated point of Z which does not belong to the closure of B, then m(θ 0 ) is a point of zero curvature of m, but m is still C 1 at θ 0 .

If θ 1 < θ 2 < θ 1 + π and [θ 1 , θ 2 ] ⊂ Z is an isolated connected component of Z, then m has a corner; i.e., it is not C 1 but has two half-tangents: a left-tangent oriented either by u(θ 1 ) if ]θ 1δ[ ⊂ F for small δ > 0, or by u(θ 1 + π) if ]θ 1δ[ ⊂ B, and a right-tangent oriented either by u(θ 2 ) or by u(θ 2 + π). Moreover, in this situation, m(α, θ 1 ) is a local center of symmetry of ∂K: the arc of ∂K in the sector ∆(α, θ 1 )

+ ∩ ∆(α, θ 2 ) -is symmetric to that in ∆(α, θ 1 ) -∩ ∆(α, θ 2 ) + .
If θ 0 is an isolated point of Z and is the endpoint of both a segment ]θ 1 , θ 0 [ of F and a segment ]θ 0 , θ 2 [ of B, then m has a cusp at m(θ 0 ). Remark. If ∂K is not C 1 then, in general, there exists α ∈ ]0, 1[ such that B(α) and F (α) are not open. Actually, if c is a corner of ∂K, with half-tangents of directions θ 1 , θ 2 , and b ∈ ∂K is such that all supporting lines to K at b have directions in ]θ 1 + π, θ 2 + π[ , then the α-section (for some α) passing through b and c has a direction θ 0 ∈ F (α) ∩ B(α), but such that any θ < θ 0 , θ close enough to θ 0 , does not belong to B(α) and any θ > θ 0 , θ close enough to θ 0 , does not belong to F (α).

Proof . (b) If ∂K is C 1 then the functions β and γ are continuous with respect to θ, hence also is v, so Z = v -1 ({0}) is a closed subset of S 1 , and

F = v -1 ( ] -∞, 0[ ) and B = v -1 ( ]0, +∞[ )
are open. Furthermore they do not intersect and their union is S 1 .

(a) Let {θ n } n∈N be a sequence converging to θ 0 , with θ n ∈ Z(α), i.e., 0 ∈ V (α, θ n ). Then a subsequence {θ n k } k∈N exists such that θ n k tends either to θ - 0 or to θ + 0 , say to θ - 0 . By [START_REF] Gutkin | Caustics for inner and outer billiards[END_REF] we have lim k→+∞ V (α, θ n k ) = v l (α, θ 0 ) , hence v l (α, θ 0 ) = 0, and θ 0 ∈ Z(α).

(c) Let θ 0 ∈ ∂F (α); then there exist sequences {θ n } n∈N in F (α) and {θ ′ n } n∈N in S 1 \F (α), both converging to θ 0 . This means that V (θ n )∩ ]0, +∞[ = ∅ and V (θ ′ n ) ⊂ ]-∞, 0] for every n ∈ N. By [START_REF] Hammer | Convex bodies associated with a convex body[END_REF], there exists

θ ′′ n ∈ [θ n , θ ′ n ] (or θ ′′ n ∈ [θ ′ n , θ n ]
) such that 0 ∈ V (θ ′′ n ) for each n, i.e., θ ′′ n ∈ Z(α). Since the sequence {θ ′′ n } n∈N tends to θ 0 and Z(α) is closed, we obtain θ 0 ∈ Z(α). The proof for ∂B(α) ⊆ Z(α) is similar. The proof for F (α) ∩ B(α) ⊆ Z(α) is obvious.

Dependence with respect to α

The functions b and c are left-and right-differentiable with respect to α; one finds, e.g., -→ ∂b ∂α l = 1 2h ( u ′ + cotanβ r u). However, we will not use their differentiability in α, but only their monotonicity. Since the function cotan is decreasing on ]0, π[ , by [START_REF] Blagojević | Convex equipartitions via equivariant obstruction theory[END_REF] we immediately obtain the following statement, whose proof is omitted. Proposition 6.1 . The functions v, v l and v r are nonincreasing in α. More precisely, we have max V (α ′ , θ) ≤ min V (α, θ) for all θ ∈ S 1 and all 0 < α < α ′ < 1.

Recall the sets F (α), B(α), and Z(α) introduced in Definition 5.1. Let

I F = {α ∈ ]0, 1[ ; F (α) = ∅}, I B = {α ∈ ]0, 1[ ; B(α) = ∅}, I Z = {α ∈ ]0, 1[ ; Z(α) = ∅}.
The symmetry [START_REF] Karasev | Equipartition of several measures[END_REF] implies

α ∈ I F ⇔ 1 -α ∈ I B and α ∈ I Z ⇔ 1 -α ∈ I Z . (17)
Set α B = inf I B and α Z = inf I Z . (We do not consider inf I F , which is always equal to 0, see below).

Theorem 6.2 . (a) We have

I B = ]α B , 1[ (and hence I F = ]0, 1 -α B [ by (17)). If ∂K is C 1 , then I Z = [α Z , 1 -α Z ]; otherwise I Z is one of the intervals [α Z , 1 -α Z ] or ]α Z , 1 -α Z [. (b) We have α Z ≤ α B ≤ 1 2 .
Moreover, if α Z = α B then ∂K contains two parallel segments.

(c) We have the following equivalences:

i. α B = 1
2 if and only if K is symmetric. ii. α Z = 0 if and only if ∂K contains a segment whose endpoints admit two parallel supporting lines to K.

iii. α B = 0 if and only if ∂K contains a segment whose endpoints admit two parallel supporting lines to K, one of which intersecting ∂K at only one point.

Remarks. 1. Notice the change of behaviour of Z and B: In Proposition 5.2, Z(α) is always closed, whereas B(α) may be not open if ∂K is not C1 ; here I B is always open, whereas I Z may be not closed if ∂K is not C 1 .

2. An example of a planar convex body such that I Z is open is the quadrilateral OICJ in Figure 9, Section 8.9, for which one finds

I Z = 1 2c , 1 -1 2c . Proof . (a) By Proposition 6.1, if 0 < α < α ′ < 1 then B(α) ⊆ B(α ′ ). It follows that I B = ]α B , 1[ or I B = [α B , 1[. We now prove that α B / ∈ I B . If α ∈ I B ,
then there exists θ ∈ S 1 such that, say, v l (α, θ) < 0 (the case v r (α, θ) < 0 is similar). Let α ′ < α and θ ′ < θ be such that b(α ′ , θ ′ ) = b(α, θ); then by (12) 1 , both v l (α ′ , θ ′ ) and v r (α ′ , θ ′ ) tend to v l (α, θ) as θ ′ → θ and α ′ → α, hence are negative for (α ′ , θ ′ ) close enough to (α, θ). This shows that α ′ ∈ I B . As a consequence, I B = ]α B , 1[ . We now prove that I Z is convex. Let α 1 < α 2 ∈ I Z ; i.e., there exist θ 1 and θ 2 in S 1 such that 0 ∈ V (α 1 , θ 1 ) ∩ V (α 2 , θ 2 ). We may assume, without loss of generality, that θ 1 ≤ θ 2 ≤ θ 1 + π. Let α ∈ ]α 1 , α 2 [. By [START_REF] Karasev | Convex equipartitions: the spicy chicken theorem[END_REF], we have V (α, θ 1 ) ∩ R -= ∅ and V (α, θ 2 ) ∩ R + = ∅ hence, by [START_REF] Hammer | Convex bodies associated with a convex body[END_REF], we have 0 ∈ V (α, θ) for some θ ∈ [θ 1 , θ 2 ] (in the case θ 2 = θ 1 + π, both intervals [θ 1 , θ 2 ] and [θ 2 , θ 1 + 2π] suit). This shows that α ∈ I Z .

(b) If α belongs neither to I F nor to I B , then we necessarily have V (α, θ) = {0} for all θ ∈ S 1 , hence the function m(α, •) is constant. This implies α = 1 2 and K symmetric. As a consequence, we have α B ≤ 1 2 . We prove 1 2 ∈ I Z , yielding

I Z = ∅. Let θ 0 ∈ S 1 arbitrary. By (7), 0 ∈ conv V 1 2 , θ 0 ∪ V 1 2 , θ 0 + π , hence by (14) 0 ∈ V 1 2 , θ for some θ ∈ [θ 0 , θ 0 + π], hence Z 1 2 = ∅. We now prove α Z ≤ α B . If α B = 1
2 , we are done; otherwise, let α ∈ I B , α ≤ 1 2 . Then v l (α, θ) or v r (α, θ) is negative for some θ ∈ S 1 , say v l (α, θ) < 0. From ( 6) and ( 16) it follows that v l (α, θ + π) = -v l (1α, θ) ≥ -v l (α, θ) > 0, hence by continuity there exists (c) ii. and iii. If α Z = 0 then there exists a sequence {α n } n∈N tending to 0, with α n ∈ I Z ; i.e., such that for any n there exists θ n ∈ S 1 with 0 ∈ V (α n , θ n ). By compactness, we may assume without loss of generality that the sequence {θ n } n∈N converges to some Conversely, if [b, c] ⊂ ∂K and b = c admit two parallel supporting lines to K, denoted by D b and D c , let θ denote the direction of (bc); i.e., θ is such that bc = k u(θ) for some k > 0. The angles

θ 1 ∈ ]θ, θ + π[ such that v l (α, θ 1 ) = 0, so θ 1 ∈ Z(α), and α ∈ I Z . If moreover α Z < α B , then consider α Z < α < α ′ < α B ≤ 1 2 , hence α, α ′ ∈ I Z \ I B . Let θ ∈ Z(α), b = b(α, θ), c = c(α, θ),
c ′ = c(α ′ , θ). Since α < α ′ , ∆(α ′ , θ) is in the interior of ∆ + (α, θ). Since B(α ′ ) = ∅,
θ 0 ∈ S 1 . Let b n = b(α n , θ n ) and c n = c(α n , θ n ).
β n = β 1 n , θ and γ n = γ 1 n , θ satisfy β n + γ n ≥ π, hence θ ∈ B 1 n ∪ Z 1 n , so B 1 n ∪ Z 1 n = ∅, 1 n ∈ I B ∪ I Z , and 0 ≤ min(α Z , α B ) = 0, hence α Z = 0 since α Z ≤ α B .
If one of the supporting lines above, say D b , intersects ∂K only at b, then b n / ∈ D b , yielding

β n + γ n > π, hence 1 n ∈ I B , showing α B = 0. Conversely, if α B = 0,
C 1 if 0 < α < α Z or 1 -α Z < α < 1, and is not C 1 if α B < α < 1 -α B . In the case α = α Z , m *
α may or may not be C 1 . However, if α Z = α B , we will see that m * α is not C 1 also for

α Z < α ≤ α B .
By Proposition 3.2 (d), K α is strictly convex (or one point or empty), and by Proposition 3.2 (e) we have m * α = ∂K α if and only if B(α) is empty. By Theorem 6.2 (a), we then have

m * α = ∂K α ⇔ α ≤ α B . (18)
Besides, the function α → K α is continuous (for the Pompeiu-Hausdorff distance on compact sets in the plane) and decreasing (with respect to inclusion): if α < α ′ , then

K α ′ K α . Since K α = ∅ if α > 1 2
, there exists a value α K ≤ 1 2 such that • K α is strictly convex with a nonempty interior if 0 < α < α K ,

• K α K is a single point denoted by T , and

• K α is empty if α K < α < 1.
By [START_REF] Klee | The critical set of a convex body[END_REF], this value α K is at least α B . It is noticeable that, when K α K is a single point, this point is not necessarily the mass center of K, see Section 8.9.

We end this section with the following statement, which gathers the last results.

Proposition 7.1 . (a) If α Z < α < 1 -α Z , then m * α is not C 1 . (b) We have 4 9 ≤ α K ≤ 1 2
, with first equality if and only if K is a triangle, and second equality if and only if K is symmetric. and[d, c] with abcd in convex position, such that the line oriented by bc, denoted by D 1 , is an α Z -section of K. Let D 2 denote the line oriented by ad. Then D 2 is a β-section for some β ≥ α B . For any α ∈ ]α Z , α B ], α > α Z but close to it, there is an α-section, denoted by ∆(α, θ 1 ), passing through c and crossing [a, b] and an α-section, denoted by ∆(α, θ 2 ), passing through b and crossing [d, c], see Figure 6. These α-sections intersect at some point P ∈ 1 2 (a + d), 1 2 (b + c) . Then we have m(α, θ) = P for all θ ∈ [θ 1 , θ 2 ], m ′ l (α, θ 1 ) collinear to P c = P c u θ 1 , and m ′ r (α, θ 2 ) collinear to bP = bP u θ 2 . Since

(c) If K is non-symmetric, then α B < α K (whereas for K symmetric we have α B = α K = 1 2 ). Proof . (a) It remains to prove that, if α Z < α ≤ α B , then m * α is not C 1 . Assume α Z < α B ;
θ 1 = θ 2 = θ 1 + π, m(α, •) is not C 1 at P . If α is not close to α Z , ∆(α, θ 1 ) (passing through c) will not cross [a, b],
and ∆(α, θ 2 ) (passing through b) will not cross [d, c]. Nevertheless, the argument given above holds as well, except that ∆(α, θ 1 ) and ∆(α, θ 2 ) will not cross on 1 2

(a + d), 1 2 (b + c . (b) Suppose α K = 1
2 . Let T be the unique point of K 1/2 . With the notation for the proof of Proposition 4.1 (a), see Figure 4, since every half-section contains T , we have

M (θ, θ ′ ) = T for all θ = θ ′ . Now this proof implies that θ -θ ′ ( b(θ) -T -c(θ) -T ) = O θ -θ ′ 2 ,
for all θ < θ ′ . Letting θ ′ tend to θ, this shows that b(θ) and c(θ) are symmetric about T .

The fact that α K ≥ 4 9 and that α K = 4 9 if and only if K is a triangle is well known; see, e.g., [START_REF] Neumann | On an invariant of plane regions and mass distributions[END_REF]. It reduces to the following statement which we prove below.

The mass center G of K is in ∆ + 4 9 , θ for all θ. Moreover, there exists θ ∈ S 1 such that ∆( 49 , θ) contains G if and only if K is a triangle. b,c) and conv(a ′′ , b, c) have an area equal to 4 9 |K|. Then the line (a ′ a ′′ ) is parallel to ∆ and for every a ∈ [a ′ , a ′′ ] the triangle conv(a, b, c) also has an area equal to 4 9 |K|. For any such a,

V = L ∪ U is in bold. sets C + = conv (∆ + ∩ D b ) ∪ (∆ + ∩ D c ) and C -= conv (∆ -∩ D b ) ∪ (∆ -∩ D c ) , one on each side of ∆. The set K + = K ∩ ∆ + is included in C + and the set K -= K ∩ ∆ -is included in C -. Let a ′ ∈ D b ∩∆ -and a ′′ ∈ D b ∩∆ -be such that the triangles conv(a ′ ,
let b ′ ∈ ∂K ∩ [a, b[ and c ′ ∈ ∂K ∩ [a, c[. By continuity, one can choose a such that the oriented line ∆ ′ = (b ′ c ′ ) is parallel to ∆. For this a, let L = conv(a, b, c). We have K -\ L ⊂ ∆ ′+ and L \ K -⊂ ∆ ′-,
therefore the y-coordinate of the mass center G -of K - is larger than or equal to the y-coordinate of the mass center of L, the equality holding only when K -is a triangle. Furthermore, since a ∈ C -, the convex set C + is included in the convex set C ′+ = conv (∆ + ∩ (ab) ∪ (∆ + ∩ (ac) .

Let U be the trapezoid of area 5 9 |K| defined by the lines (ab), (ac), ∆ and a line ∆ ′′ ⊂ ∆ + parallel to ∆. Since K + \ U ⊂ ∆ ′′+ and U \ K + ⊂ ∆ ′′-, the y-coordinate of the mass center G + of K + is larger than or equal to the y-coordinate of the mass center of U , the equality holding only when K + = U . It follows that G y is larger than or equal to the y-coordinate of the mass center of the union V of L and U , with equality only if K = V . Since the mass center of V is on ∆ we are done.

(c) Suppose that α B = α K . By [START_REF] Klee | The critical set of a convex body[END_REF], we obtain that m(α K , •) is constant and equal to T . It follows that, for all θ ∈ S 1 , T is the middle of a chord of direction θ, hence K is symmetric about T .

8 Miscellaneous results, remarks, and questions 8.1 It is easy to check that, if an α-section crosses two non-parallel segments of ∂K, then the corresponding middle of chord m(α) lies on an arc of hyperbola asymptotic to the lines extending these segments. In particular, if K is a convex polygon, then for all α ∈ ]0, 1[ the curve m(α, •) is entirely made of arcs of hyperbolae. Of course, not only segments yield arcs of hyperbolae. One can check for instance that two arcs of hyperbolae on ∂K also give an arc of hyperbola for m(α, •), if α is small enough.

8.2

We saw that an envelope of α-sections is a tour in the sense of Definition 3.1. Conversely, which tours are envelopes of some α-sections m(α, K), for which convex bodies K and for which 0 < α < 1? For example, the astroid on the left of Figure 2 given by θ → (-cos 3 θ, sin 3 θ), i.e., with R(θ) = 3 2 sin(2θ), cannot be such an envelop. More generally, following a remark by D. Fuchs and S. Tabachnikov in [START_REF] Fuchs | Mathematical omnibus: Thirty lectures on classic mathematics[END_REF], a tour with a common tangent at two different points cannot be such an envelop, since each point would have to be the middle of the chord.

8.3

The KAM theory applied to dual billiards shows that, if m is a tour of class C 5 (hence strictly convex), then there exist convex bodies K with ∂K arbitrarily close to m such that m = m(α, K) (hence for some α arbitrarily close to 0) and there exist convex bodies K with ∂K arbitrarily close to infinity such that m = m(α, K) (hence for some α arbitrarily close to 1 2 ). The curves ∂K are invariant torii of the dual billiard. According to E. Gutkin et A. Katok [START_REF] Gutkin | Caustics for inner and outer billiards[END_REF], the works of J. Moser [START_REF] Menon | A theorem on partitions of mass-distribution[END_REF] and R. Douady [START_REF] Day | Polygons circumscribed about closed convex curves[END_REF] prove that these curves are convex. Notice the following apparent paradox. By Theorem 6.2(c)i., if K is non-symmetric then, for α close to 1 2 , m(α) is non-convex. This apparently contradicts the above results which imply that, given a strictly convex C 5 non-symmetric curve, there exist values of α arbitrarily close to 1 2 and convex bodies K, necessarily non-symmetric, such that m = m(α, K). However, these curves have envelops m(β, K) with cusps for some other β ∈ ]α, 1 2 [, although α is arbitrarily close to 1 2 . There is no real contradiction. Several questions remain open: Among tours which present cusps, which ones are envelopes of sections? For instance, does it exist a symmetric curve with cusps which is the α-envelope of some (necessarily non-symmetric) convex body? Also here, does it exist a non-symmetric convex body K having a symmetric convex envelope m(K, α, •) for some α ∈ ]0, 1 [ ? 8.4 The link between envelopes of α-sections and dual billiards yields a simple proof for the following (also simple) fact:

The only convex bodies which have a circular envelope (or an elliptic one, since this is the same question modulo an affine transformation) are the discs having the same centers as the envelop. Actually, if the billiard table is a circle, then the billiard map is integrable 2 . This means that each orbit remains on a circle centered at the origin, and is either periodic or dense in this circle, depending whether the angle between the two tangents from the starting point at the table is in πQ or not. If the convex body K were not a disc, then its boundary would cross at least one circle with dense orbits, hence would contain at least one dense orbit on the latter circle, hence the whole circle, a contradiction.

8.5

Let C be a closed subset of S 1 . We construct below a C 1 convex curve which is the boundary of a convex body for which Z 1 2 = C ∪ S, for some countable set S. As convex curve, we start with the unit circle and will deform it. Since C is closed, S 1 \ C is a countable union of open intervals. Let ]θ 1 , θ 2 [ be one of them. Between θ 1 and θ 2 , we deform the circle into a convex C 1 curve arbitrarily close to the union of two segments, one [e iθ 2 , p] tangent to the circle at e iθ 2 , the other [e iθ 1 , p], such that the area remains unchanged. (For convenience, here we use the notation of complex numbers.) We do the same symmetrically with respect to the line passing through the origin and of direction 1 2 (θ 1 + θ 2 + π); i.e., we choose a C 1 curve close to [e i(θ 1 +π) , p ′ ] tangent to the circle at e i(θ 1 +π) , and to [e i(θ 2 +π) , p ′ ], with p ′ = p e i(θ 1 +θ 2 +π) , see Figure 8.

Observe that the curve is no longer (centrally-)symmetric. If the curve is chosen with large curvature near the points p and p ′ , then the interval ]θ 1 , θ 2 [ contains only two values of θ for which the half-section cuts the curve in a chord with two parallel tangents; these chords have one endpoint near p, resp. near p ′ , and do not contain the origin anymore.

Doing this in any connected component of S 1 \ C, we obtain a convex C 1 curve such that any θ ∈ C has a halving chord with parallel tangents and all but two values of θ in each connected component of S 1 \ C have a halving chord without parallel tangents. It is easy to construct K L such that L α ⊂ K α for some value of α; for example, L is an equilateral triangle, K the inscribed circle of L, and α ∈ 4 9 , 1 2 . Our calculation shows that this holds for all α ∈ α 1 , 1 2 , where α 1 ≈ 0.40716 satisfies 1

-3 2 (1 - √ 1 -2α 1 ) = cos t and α 1 = 1 π (t -cos t sin t). Actually, if the radius of K is R (hence the height of L is 3R), K α is a disc of radius R α = R cos t such that α = 1 π (t - cos t sin t).
Then ∂L α is made of three arcs of hyperbolae. In a (non-orthonormal) frame where the triangle has vertices (0, 0), (0, 1), and (1, 0), two of these arcs of hyperbolae are xy = α 4 and x(1xy) = α 4 , hence they cross at a point of abscissa

x c = 1 2 (1 - √ 1 - 2α
)) This proves that the three arcs of ∂L α cross at a distance R - 3 2 R(1 -√ 1 -2α) from the center.

Is it true that, for every value of α ∈ ]0, 1 2 [ there exists K L such that L α ⊂ K α ? (It cannot exist K, L, K L, independent of α such that L α ⊂ K α for all small α > 0).

On the contrary, does it exist α 1 > 0 such that for all pairs K, L, K L, and all α < α 1 , L α ⊂ K α ? Is α 1 ≈ 0.40716 optimal with this respect? I.e., is it optimal when K is a disc and L a triangle? Or, instead, is it optimal when K is an affinely regular hexagon inscribed in L, a triangle? 8.7 The following questions have been asked by Jin-ichi Itoh, whom we would like to thank for his interest in our work. Let w(K) denote the width of a convex body K, ⊘(K) its diameter, r(K) its inradius, and R(K) its circumradius.

Do we have w(K)

⊘(K) ≤ w(Kα) ⊘(Kα) for all α < 1 2 ?

Do we have r(K) R(K) ≤ r(Kα) R(Kα) for all α < 1 2 ? Both answers are "no", as shown by the following counter-examples. For the first question, if K is the unit Reuleaux triangle, ⊘(K) = w(K) = 1, and α > 0 is small enough, then one can check that ⊘(

K α ) = 1 -O α 2/3 , whereas w(K α ) < 1 -α 1/2 < ⊘(K α ).
For the second question, consider two small arcs of the circle of center 0 and radius 2 near the x-axis, two small arcs of the circle of center 0 and radius 1 near the y-axis, and choose for K the convex hull of the union of these four arcs.

In this manner we have r(K) R(K) = 1 2 . However, we have r(K α ) = cos t 1 , where t 1 is such that α

= 1 |K| (t 1 -sin t 1 cos t 1 ) = 2 3|K| t 3 1 + O(t 5
α = 4 |K| (t 2 -sin t 2 cos t 2 ) = 8 3|K| t 3 2 + O(t 5
2 ), hence t 2 < t 1 , so r(Kα) R(Kα) < 1 2 . Whether there is a constant k > 0 such that w(K) ⊘(K) ≤ k w(Kα) ⊘(Kα) for all convex bodies K and all α, idem for r(K) R(K) , and which constant is optimal, seems to be another interesting question.

8.8

There is a tight link between the region where appear cusps, as described in [START_REF] Fuchs | Mathematical omnibus: Thirty lectures on classic mathematics[END_REF], and the regions M 2 and M 3 described by T. Zamfirescu in [START_REF] Zamfirescu | Spreads[END_REF]. Actually, let K be a planar convex body and, for each θ ∈ S 1 , consider the so-called midcurve, i.e., the locus of midpoints of all chords of direction θ. If ∂K is C 1 and strictly convex then the family of all these midcurves for all θ ∈ S 1 is a continuous family in the sense of Grünbaum [START_REF] Grünbaum | Continuous families of curves[END_REF], for which general results of [START_REF] Zamfirescu | Spreads[END_REF] are at our disposal.

In our situation, the regions M 2 and M 3 of [START_REF] Zamfirescu | Spreads[END_REF] are the loci of points of Int (K) which are middles of at least 2, resp. 3, different chords. If K is symmetric then M 2 = M 3 = {g(K)}, the mass center of K. If K is not symmetric then it seems that we have

Int (M 2 ) = M 3 = Int α∈ ]0, 1 2 [ (m * α \ ∂K α ) . (19)
Already ( 19) can be verified when K is a quadrilateral. For instance, the polygonal curve of all cusps described in Figure 11.15 of [START_REF] Fuchs | Mathematical omnibus: Thirty lectures on classic mathematics[END_REF] is indeed the boundary of M 3 . Besides, when K is a quadrilateral, a detailed analysis shows that the unique point T of K α K is in M 3 and that the three chords bisected by T are all α K -sections. This property of T is likely to be true in the general case. Conversely, is T the only point of M 3 bisecting three α-sections with the same α? This has been checked in the case of quadrilaterals.

8.9

We present here an explicit example showing that the unique point T of K α K is not necessarily the mass center G of K. (the midpoint of the segment determined by the centers of mass of the triangles OJC and OIC). The point G is the midpoint of exactly three chords, AB, EF , and E ′ F ′ , with A, B on the line of equation x + y = 2c+1

3 , E ∈ OJ, F ∈ IC, and E ′ ∈ OI, F ′ ∈ JC, see Figure 9. These chords divide K in 2 ). By symmetry, we have α EF = α E ′ F ′ , but there is no reason that α EF = α AB ; as we will see, this is indeed not the case.

E T ¨¨¨¨¨¨¨¨¨¨¨¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ 4
The triangle ABC is the image of IJC by the dilation of center C sending the point 6 + 4c-1 9 = 8c+1 18 . This gives α EF = 8c+1 18c = α AB , hence G = T . More generally, given a planar convex body G, we consider all chords the midpoint of which coincides to the mass center G of K. It is known [START_REF] Viet | Umkehrung eines Satzes von H. Brunn über Mittelpunktseibereiche[END_REF] that there are at least three different such chords; these chords are α-sections for different values of α ranging from some α min (K) to some α max (K). Then the quotient α min (K) αmax (K) measures in some sense the asymmetry of the body K. It is equal to 1 if K is (centrally) symmetric, or if K has another symmetry which ensures that G is the only affine-invariant point, e.g. for K a regular polygon, but it differs from 1 in general. One can see that the minimum of this quotient is achieved for at least one affine class of convex bodies, and it would be worthy to determine the shape of these bodies, and to compute the corresponding minimum. In the case of all quadrilaterals, using Maple, we found that the minimum is attained precisely for a quadrilateral of our previous family, for c = 7 4 and with α min (K) αmax (K) = 24 25 . The same questions can be asked in higher dimensions, replacing the midpoint of a chord by the mass center of an α-section.

H = 1 2 , 1 2 to G,

8.10

We end this section with our main conjecture. Conjecture 8.1 . For any convex bodies K, L with K ⊂ L, and any α ∈ ]0, 1 2 [ , there exists an α-section of L which is a β-section of K for some β ≤ α.

This conjecture has been recently proven in [START_REF] Fruchard | Fair partitioning by straight lines[END_REF] in the case of planar convex bodies. Another natural related question, the answer of which turns out to be negative, is the following.

For K ⊂ L convex bodies, does there always exist a half-section ∆ of L such that |K| length(∆ ∩ L) ≤ |L| length(∆ ∩ K) ? For a counter-example, consider for K a thin pentagon of width varying from ε on each side to 2ε in the middle, placed at the basis of L, a triangle of heigth 1, see Figure 10. A t t t t t t t t t t t @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @h h

h h h h h h h h h h h h h h h h h h h t t t t t t

  e., b ′ l = b ′ r ); otherwise we call b a corner point of ∂K (a corner for short). If b is regular then β = ( b ′ , u) ∈ ]0, π[ denotes the angle between the tangent to ∂K at b and ∆. If b is a corner then β l = ( b ′ l , u), resp. β r = ( b ′ r , u), is the angle between the left-tangent, resp. right-tangent, to ∂K at b and ∆. Similarly, let γ

Figure 2 :Proposition 3 . 2 .

 232 Figure2: Some (double half-)tours: Left the astroid, right the deltoid. The astroid cannot be an envelope of an α-section of a planar convex body; the deltoid is probably such a half-section, although we ignore how to prove it.

  (b) If R l and R r do not vanish, then s is a homeomorphism (with inverse homeomorphism denoted θ for convenience) and m θ(s) = m(0) + s 0 u θ(t) dt, with u and θ continuous, i.e., m • θ is of class C 1 .

Lemma 3 . 3 .

 33 The interior of the core K of m coincides with the intersection of all open half-planes Int D + (θ) , θ ∈ S 1 :

Figure 3 :

 3 Figure 3: Proof of Lemma 3.4 and Proposition 3.2 (d)

Proposition 4 . 1 .

 41 (a) The curve m is the envelope of the family {∆(α, θ), θ ∈ S 1 }, i.e., the vector products m ′ l ∧ u and m ′ r ∧ u vanish identically. (b) If b and c are regular points of ∂K then

  If b and c are regular then v is the signed radius of curvature of the curve m. If b or c (or both) is a corner then v l , resp. v r , is the signed radius of curvature on the left, resp. on the right, of m.

Figure 4 :

 4 Figure 4: Proof of Proposition 4.1

Proposition 5 . 2 .

 52 Let α ∈ ]0, 1[. (a) The set Z(α) is closed in S 1 . (b) If ∂K is C 1 then the sets F (α) and B(α) are open in S 1 and F (α), B(α), and Z(α) form a partition of S 1 . (c) In the general case, the three sets F (α) ∩ B(α), ∂F (α), and ∂B(α) are subsets of Z(α).

  and let D b , D c be two parallel supporting lines to K at b and c respectively, see Figure 5. Consider now b ′ = b(α ′ , θ) (for the same θ) and

Figure 5 :

 5 Figure 5: Proof of Theorem 6.2 (b)

  there do not exist supporting lines to K at b ′ and c ′ crossing in ∆ + (α ′ , θ). As a consequence, b ′ must lie on D b and c ′ on D c , and the segments [b ′ , b] and [c ′ , c] are on ∂K. (c) i. If α B = 1 2 then by (a) 1 2 / ∈ I F ∪ I B hence, as already said in the proof of (b), K is symmetric. The converse is obvious.

  By continuity, the sequences {b n } n∈N and {c n } n∈N converge to some b, resp. c ∈ ∂K. Since the width ofK satisfies 0 < w(K) ≤ b nc n , we have b = c. Since 0 ∈ V (α n , θ n ),there are parallel supporting lines to K at b n and c n for each n. Without loss of generality, we also assume that these lines converge, yielding two parallel supporting lines at b and c, denoted D b and D c . Since the sequence of lines ∆(α n , θ n ) n∈N tends to the oriented line through b and c, denoted by (bc), and |∆ -(α n , θ n ) ∩ K| = α n |K| → 0, we obtain |(bc) -∩ K| = 0, hence the segment [b, c] is on ∂K.

  then the former points b n and c n admit supporting lines which cross in ∆ + (α n , θ n ) (see comment after Definition 5.1). By contradiction, if both D b ∩ ∂K and D c ∩ ∂K contained more than b, resp. c, then for n large enough we would have b n ∈ D b and c n ∈ D c . Then for m, n such that b m ∈ ]b n , b[ and c m ∈ ]c n , c[ , the only supporting lines at b m , c m could be D b and D c , which do not cross, a contradiction. 7 The α-core In this section we compare the boundary ∂K α with the image of m(α, •), denoted by m * α , for α ∈ ]0, 1[ . The function of θ, m(α, •), is a tour in the sense of Definition 3.1, with R l = v l (α, •) and R r = v r (α, •), and core K α . In the case α = 1 2 , m 1 2 , • is also a double half-tour. Then, by Proposition 3.2 (b) and (c) and Theorem 6.2 (a), m * α is of class

  by Theorem 6.2 (b) and its proof, there exist two parallel segments on ∂K, denoted by[a, b]

Figure 6 :

 6 Figure 6: Proof of Proposition 7.1 (a)

Figure 7 :

 7 Figure 7: Proof of Proposition 7.1 (b). The triangle V = L ∪ U is in bold.

Figure 8 :

 8 Figure 8: A construction of C 1 convex body with (almost) prescribed zero set

  The strategy of proof is to use a quadrilateral and to show that the three chords bisected by G are α-sections for different values of α, showing that T = G by Subsection 8.8. Given c > 1, let K = OIJC denote the quadrilateral conv(O, I, J, C), with O = (0, 0), I = (1, 0), J = (0, 1), C = (c, c). The area of K is |K| = 1 2 -→ IJ -→ OC = c and its mass center is G = 2c+1 6 , 2c+1 6

Figure 9 :

 9 Figure 9:The α-center is not the mass center; here c = 2proportion (α AB , 1α AB ), resp. (α EF , 1α EF ) and (α E ′ F ′ , 1α E ′ F ′ ) (by convention, 0 < α XY ≤12 ). By symmetry, we have α EF = α E ′ F ′ , but there is no reason that α EF = α AB ; as we will see, this is indeed not the case.The triangle ABC is the image of IJC by the dilation of center C sending the point

  -1) . With |IJC| = c -1 2 , this gives|ABC| = (4c-1) 2 18(2c-1) , so α AB = (4c-1) 2 18c(2c-1) . Since F is on the line IC of equation c(x -1) = (c -1)y, one finds F = 2c+1 3 , 2c 3 , hence E = 0, 1 3 . The area of the triangle EIF is 1 2 det( of the quadrilateral OIF E is |OIF E| = |OIE| + |EIF | = 1

Figure 10 :

 10 Figure 10: In bold, the body K; in thin, the body L; here ε = 1 10 detailed analysis shows that, for every half-section ∆ of L, one has |K| length(∆ ∩ L) > |L| length(∆ ∩ K).

  1 ), and R(K α ) = 2 cos t 2 , where t 2 is such that

In fact[START_REF] Grünbaum | Continuous families of curves[END_REF] is stated for α fixed, but its proof uses only (3) and (10) -which can easily be adapted to our situation -and the continuity of the function cotan.

A classical conjecture states that only ellipses have a dual billiard map which is integrable.

Acknowledgements. The authors thank Theodor Hangan and Tudor Zamfirescu for fruitful discussions and for having pointed to them several references. They are also indebted to the anonymous referee who drew to their attention the reference [21]. The third author thanks the Universitï¿ 1 2 de Haute Alsace for a one-month grant and its hospitality, and acknowledges partial support from the Romanian National Authority for Scientific Research, CNCS-UEFISCDI, grant PN-II-ID-PCE-2011-3-0533.

Nicolas Chevallier and Augustin Fruchard Laboratoire de Mathï¿