Two Triangles in a Circle

Let $[ABC]$ be the area of $\triangle ABC,$ and define $[XYZ]$ similarly. It can be shown that

$16[XYZ]^3 \geq 27R^4[ABC],$

where $R$ is the radius of $\omega$ i.e. the circumcircle of $\triangle ABC.$ Since $R = 1,$ we have $16[XYZ]^3 \geq 27[ABC],$ or $[XYZ]^3 \geq \dfrac{27}{16} [ABC].$ Thus, $p + q = 27 + 16 = \boxed{43}.$

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In this section, we shall prove the inequality. (This is probably not the best way to prove it, since I'm still learning how to tackle inequalities. Let me know if there are any errors in the following proof.)

Recall that the area of $\triangle ABC$ is equal to $2R^2 \sin A \sin B \sin C.$ To find the angle measures of $\triangle XYZ,$ we see that

$\angle X = \angle AXZ + \angle AXY = \angle ACZ + \angle ABY = \dfrac{1}{2} (\angle B + \angle C).$

Similarly, $\angle Y = \dfrac{1}{2} (\angle C + \angle A)$ and $\angle Z = \dfrac{1}{2} (\angle A + \angle B).$ Thus, the area of $\triangle XYZ$ can be written as $2R^2 \sin \dfrac{B+C}{2} \sin \dfrac{C+A}{2} \sin \dfrac{A+B}{2}.$ We can manipulate this to get

$\begin{aligned} [XYZ] &= 2R^2 \sin \dfrac{B+C}{2} \sin \dfrac{C+A}{2} \sin \dfrac{A+B}{2} \\ &= 2R^2 \sin \left ( 90 - \dfrac{A}{2} \right ) \sin \left ( 90 - \dfrac{B}{2} \right ) \sin \left ( 90 - \dfrac{C}{2} \right ) \\ &= 2R^2 \cos \dfrac{A}{2} \cos \dfrac{B}{2} \cos \dfrac{C}{2}. \end{aligned}$

Plugging these values in to our given inequality gives us

$\begin{aligned} 16[XYZ]^3 &\geq 27R^4[ABC] \\ 16 \left ( 2R^2 \cos \dfrac{A}{2} \cos \dfrac{B}{2} \cos \dfrac{C}{2} \right )^3 &\geq 27R^4(2R^2 \sin A \sin B \sin C) \\ 64 \cos^3 \dfrac{A}{2} \cos^3 \dfrac{B}{2} \cos^3 \dfrac{C}{2} &\geq 27 \sin A \sin B \sin C \\ 64 \cos^3 \dfrac{A}{2} \cos^3 \dfrac{B}{2} \cos^3 \dfrac{C}{2} &\geq 27 \left (2 \sin \dfrac{A}{2} \cos \dfrac{A}{2} \right ) \left (2 \sin \dfrac{B}{2} \cos \dfrac{B}{2} \right ) \left (2 \sin \dfrac{C}{2} \cos \dfrac{C}{2} \right ) \\ 8 \cos^2 \dfrac{A}{2} \cos^2 \dfrac{B}{2} \cos^2 \dfrac{C}{2} &\geq 27 \sin \dfrac{A}{2} \sin \dfrac{B}{2} \sin \dfrac{C}{2} \\ 8 \left (1 - \sin^2 \dfrac{A}{2} \right ) \left (1 - \sin^2 \dfrac{B}{2} \right ) \left (1 - \sin^2 \dfrac{C}{2} \right ) &\geq 27 \sin \dfrac{A}{2} \sin \dfrac{B}{2} \sin \dfrac{C}{2}. \end{aligned}$

Let $a = \sin \dfrac{A}{2}, b = \sin \dfrac{B}{2}, c = \sin \dfrac{C}{2}.$ Our inequality is equivalent to $8(1 - a^2)(1 - b^2)(1 - c^2) \geq 27abc.$

Since $0^{\circ} < A, B, C < 180^{\circ},$ we see that $0 < a, b, c < 1,$ so $1 - a^2, 1 - b^2, 1 - c^2 > 0.$ By the AM-GM Inequality, we have

$\dfrac{abc}{(1 - a^2)(1 - b^2)(1 - c^2)} \leq \left ( \dfrac{\frac{a}{1 - a^2} + \frac{b}{1 - b^2} + \frac{c}{1 - c^2}}{3} \right )^3.$

Next, applying Titu's Lemma, we get

$\begin{aligned} \dfrac{a}{1 - a^2} + \dfrac{b}{1 - b^2} + \dfrac{c}{1 - c^2} &= \dfrac{1}{\frac{1}{a} - a} + \dfrac{1}{\frac{1}{b} - b} + \dfrac{1}{\frac{1}{c} - c} \\ &\leq \dfrac{9}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} - (a + b + c)}. \end{aligned}$

Let $k = \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}.$ By Cauchy-Schwarz, $a + b + c \leq \dfrac{9}{k},$ so

$\begin{aligned} \dfrac{9}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} - (a + b + c)} &\leq \dfrac{9}{k - \frac{9}{k}} \\ &= \dfrac{9k}{k^2 - 9}. \end{aligned}$

Thus, we must determine if $\dfrac{9k}{k^2 - 9} \leq 2.$ Indeed, this inequality is equivalent to $2k^2 - 9k - 18 \geq 0,$ or $(2k + 3)(k - 6) \geq 0.$ But, we know that $k \geq 6,$ which can be proved by applying the Erdös-Mordell Inequality for the incenter of a triangle. The desired result follows. Equality occurs when $\triangle ABC$ is equilateral. $\blacksquare$