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A cyclic hexagon is inscribed inside a circle. The sum of two consecutive sides always equals 149. Then, we triangulate the hexagon into four triangles each containing an incircle, and surprisingly, the sum of the four inradii equals the radius of the large circumradius. What’s the length of the hexagon’s smallest side?

My attempt: I tried to solve using the inradius and the circumradius theorems, but there will be too much calculation. Hence, I want a new way to solve it.

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According to the Japanese theorem for cyclic polygons, the sum of incircle radii does not change no matter how you triangulate this inscribed hexagon.

So let's triangulate the hexagon differently so we can use its symmetry:

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Let $a = |BC|, b = |AC|, c = |AB|$, wlog. let $a > c$. Let $R$ be the exradius, and $r$ be the inradius of the triangle $ABC$. The inradius of the equilateral triangle $ACE$ is equal to $R/2$, so the sum of the four inradii is $R/2 + r + r + r = R$. Therefore $r = R/6$.

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Applying Carnot's theorem to the triangle ABC yields $$R + r = |M_{AB}M| + |M_{BC}M| - |M_{AC}M|.$$ Using that $r = R/6$, $|M_{AC}M| = R/2$ and $|M_{BC}M| = |M_{DE}M|$ by symmetry, we get $$|M_{AB}M_{DE}| = |M_{AB}M| + |M_{DE}M| = R + R/2 + R/6 = \frac{5}{3}R.$$

All angles in the hexagon are equal by symmetry, so they are equal to $120^\circ$. Note now that $|M_{AB}M_{DE}|$ is the altitude of a regular triangle $ADG$ with side length $a+c$, so we have $\frac{\sqrt{3}}{2}(a+c) = \frac{5}{3}R$.

Also note that $\frac{3}{2}R$ is the altitude of the regular triangle $ACE$ with side length $b$, so we have $\frac{3}{2}R = \frac{\sqrt{3}}{2}b.$

Together, the last two equalities yield $b = \frac{9}{10}(a+c) = \frac{9\cdot 149}{10}$.

Now the cosine rule in the triangle $ABC$ says $b^2 = a^2+ac+c^2 = (a+c)^2 - ac$, so we get $$\left(\frac{9\cdot 149}{10}\right)^2 = 149^2 - ac$$. Simplifying yields $$ac = \frac{19\cdot 149^2}{100}$$.

Now we know the sum and product of $a$ and $c$, therefore we can obtain them as solutions $x$ of the equation $0 = (x-a)(x-c) = x^2-(a+c)x + ac = x^2 - 149x + \frac{19\cdot 149^2}{100}$. The solutions of this quadratic equation are $$149 \cdot \frac{5\pm\sqrt{6}}{10}.$$ So the length of the smallest side of the hexagon, $c$, is equal to $149 \cdot \frac{5-\sqrt{6}}{10}$.

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  • $\begingroup$ Ah, I had the same initial idea but then I decided to use the formula for inradius in terms of half-angles and got myself in some deep muck. Elegant solution! $\endgroup$ – AxiomaticSystem May 27 at 15:11
  • $\begingroup$ What an elegant solution! Very impressive! $\endgroup$ – user69590 May 27 at 15:21

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