Newsletter #45 - Solution

Problem

Let <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>M</mi></mrow><annotation encoding="application/x-tex">M</annotation></semantics></math>M be the largest distance among six distinct points of the plane, and let <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>m</mi></mrow><annotation encoding="application/x-tex">m</annotation></semantics></math>m be the smallest of their mutual distances. Prove that

Solution

Let <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>A</mi><mi>B</mi><mi>C</mi></mrow><annotation encoding="application/x-tex">ABC</annotation></semantics></math>ABC be a triangle with sides <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>a</mi><mo>=</mo><mi>B</mi><mi>C</mi><mo>≥</mo><mi>b</mi><mo>=</mo><mi>C</mi><mi>A</mi><mo>≥</mo><mi>c</mi><mo>=</mo><mi>A</mi><mi>B</mi></mrow><annotation encoding="application/x-tex">a = BC \geq b = CA \geq c = AB</annotation></semantics></math>a=BC≥b=CA≥c=AB. Suppose that angle <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>A</mi><mo>≥</mo><mfrac><mrow><mn>2</mn><mi>π</mi></mrow><mn>3</mn></mfrac></mrow><annotation encoding="application/x-tex">A \geq \frac{2\pi}{3}</annotation></semantics></math>A≥32π. It follows immediately from the cosine formula we have <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msup><mi>a</mi><mn>2</mn></msup><mo>=</mo><msup><mi>b</mi><mn>2</mn></msup><mo>+</mo><msup><mi>c</mi><mn>2</mn></msup><mo>−</mo><mn>2</mn><mi>b</mi><mi>c</mi><mi>cos</mi><mo>⁡</mo><mi>A</mi><mo>≥</mo><msup><mi>b</mi><mn>2</mn></msup><mo>+</mo><msup><mi>c</mi><mn>2</mn></msup><mo>+</mo><mi>b</mi><mi>c</mi><mo>≥</mo><mn>3</mn><msup><mi>c</mi><mn>2</mn></msup></mrow><annotation encoding="application/x-tex">a^2 = b^2 + c^2 - 2bc \cos A \geq b^2 + c^2 + bc \geq 3c^2</annotation></semantics></math>a2=b2+c2−2bccosA≥b2+c2+bc≥3c2, and hence that <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mfrac><mi>a</mi><mi>c</mi></mfrac><mo>≥</mo><msqrt><mn>3</mn></msqrt></mrow><annotation encoding="application/x-tex">\frac{a}{c} \geq \sqrt{3}</annotation></semantics></math>ca≥3<svg xmlns="http://www.w3.org/2000/svg" width="400em" height="1.08em" viewBox="0 0 400000 1080" preserveAspectRatio="xMinYMin slice"><path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"/></svg>. So it is sufficient to find a triangle with an angle of at least <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mfrac><mrow><mn>2</mn><mi>π</mi></mrow><mn>3</mn></mfrac></mrow><annotation encoding="application/x-tex">\frac{2\pi}{3}</annotation></semantics></math>32π.

If the <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>6</mn></mrow><annotation encoding="application/x-tex">6</annotation></semantics></math>6 points form a convex hexagon, then the six angles of the hexagon sum to <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>4</mn><mi>π</mi></mrow><annotation encoding="application/x-tex">4\pi</annotation></semantics></math>4π so at least one is at least <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mfrac><mrow><mn>2</mn><mi>π</mi></mrow><mn>3</mn></mfrac></mrow><annotation encoding="application/x-tex">\frac{2\pi}{3}</annotation></semantics></math>32π. If not then one point is inside the convex hull of the others. Draw diagonals to triangulate the hull. Then the inside point must lie inside (or on) one of the triangles. But if <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>P</mi></mrow><annotation encoding="application/x-tex">P</annotation></semantics></math>P lies inside (or on) the triangle <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>A</mi><mi>B</mi><mi>C</mi></mrow><annotation encoding="application/x-tex">ABC</annotation></semantics></math>ABC, then at least one of the angles <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>A</mi><mi>P</mi><mi>B</mi></mrow><annotation encoding="application/x-tex">APB</annotation></semantics></math>APB, <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>B</mi><mi>P</mi><mi>C</mi></mrow><annotation encoding="application/x-tex">BPC</annotation></semantics></math>BPC, <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>C</mi><mi>P</mi><mi>A</mi></mrow><annotation encoding="application/x-tex">CPA</annotation></semantics></math>CPA is at least<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mfrac><mrow><mn>2</mn><mi>π</mi></mrow><mn>3</mn></mfrac></mrow><annotation encoding="application/x-tex">\frac{2\pi}{3}</annotation></semantics></math>32π.

Thank you!

Thanks to Charles Burns (Oxford) and Chris (Italy) for submitting solutions to this week's challenge problem.