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Putnam 1967 B1 "A convex hexagon is inscribed in a circle. The length of every other edge equals the radius of the circle. Prove that the triangle formed by the midpoints of the remaining three edges is equilateral." Create a figure conforming to the conditions of the problem. Merely describe the sequence of angles counterclockwise starting at (1, 0) and compute their coordinates: that's the figure. In[35]:= Clear@hexplotD; hexplot@Α_, Β_D := With@ 8Γ = Π - HΑ + ΒL, Ω = Π 3<, H* Α, Β, and Γ measure the irregular arcs in the hexagon. *L 8Cos@ðD, Sin@ðD< & Accumulate@80, Α, Ω, Β, Ω, Γ, Ω<D H* Output is a closed list of vertices, starting and ending at H1,0L. *L D Connect every other midpoint. Use the midpoint formula: midpoints are averages of the endpoints. In[37]:= Clear@midpointD; midpoint@data_D := HðP1T + ðP2TL 2 & Partition@data, 2D ; Close a figure. Tack the first point onto the end of a list of points to close it up. In[39]:= Clear@closeD; close@s_D := Append@s, Part@s, 1DD; Draw a picture. Let the user choose Α and Β, then draw the circle, the hexagon, and the triangle of midpoints: all the hard work was done above. Printed by Mathematica for Students 2 3. Hexagon.nb In[41]:= Manipulate@ Module@ 8data = hexplot@Α, ΒD, unitCircle = ParametricPlot@8Cos@ΘD, Sin@ΘD<, 8Θ, 0, 2 Π<, PlotStyle ® 8Red<D <, hexagon = ListPlot@data, Joined ® True, PlotStyle ® 8Thick<D; vertices = ListPlot@data, PlotStyle ® 8PointSize@LargeD<D; midpoints = ListPlot@close@midpoint@dataDD, Joined ® True, 8PlotStyle ® 8Thick, Gray, Dashed<<D; Show@unitCircle, hexagon, vertices, midpoints, Axes ® FalseD D, H* Allowing negative angles is illuminating! *L 88Α, Π 3<, - Π, Π<, 88Β, Π 3<, - Π, Π< D Α Β Out[41]= Solve the problem by showing that one angle of the triangle equals 2Π/3. This solution exploits complex arithmetic, so create a function to convert points into their complex equivalents. In[42]:= Clear@complexD; complex@8x_, y_<D := x + ä y; Compute the sides of the midpoint triangle. Printed by Mathematica for Students 3. Hexagon.nb In[44]:= 3 Clear@Α, ΒD; sides = complex HDifferences@close@midpoint@hexplot@Α, ΒDDDDL; sides MatrixForm Out[46]//MatrixForm= 1 H- 1 - Cos@ΑDL + ä J 2 äK 1 2 J- CosB Π 6 - ΑF - CosB 1 2 In[47]:= Π 6 1 JCosB 2 - Α - ΒFN + 1 2 6 - ΑF + CosB 3 K- H1 + Cos@ΑDL + ä K Π 2 1 2 + CosB K 3 2 Π 6 Π 6 - Α - ΒFN - + Α + ΒFOO + - CosB Π 6 1 2 + Α + ΒFO + Sin@ΑD 2 N+ J- SinB Sin@ΑD 2 Π 6 O+ 1 2 JSinB Π 6 - ΑF + SinB - ΑF - SinB 1 2 J- 1 2 Π 6 + SinB Π 6 Out[48]//MatrixForm= 4 1 4 1 4 J1 + ä J1 - ä J- 2 + ãä Α J- 1 + ä 3 + J1 + ä 3 - ãä Α - 2 ãä HΑ+ΒL + 3 + 2 Cos@ΑD - 2 ä CosB Π 6 3 N ãä Β NN 3 H- ä Cos@ΑD + Sin@ΑDLN + Α + ΒF + 2 ä Sin@ΑD + 2 SinB Π 6 + Α + ΒFN Find the angle between the first two sides. In[49]:= Arg@sidesP2T sidesP1TD FullSimplify 2Π Out[49]= 3 Because these sides were arbitrary, each one is obtained by rotating another by 2Π/3: it is an equilateral triangle. Printed by Mathematica for Students 6 - Α - ΒF - Α - ΒFN + sides = sides FullSimplify; sides MatrixForm 1 Π + Α + ΒFN 1 2 J 1 2 -