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2018 MPFG Problem 10

Revision as of 07:25, 8 November 2025 by Cassphe (talk | contribs) (→Solution 1)

Problem

Let $T_1$ be an isosceles triangle with sides of length $8$, $11$, and $11$. Let $T_2$ be an isosceles triangle with sides of length $b$, $1$, and $1$. Suppose that the radius of the incircle of $T_1$ divided by the radius of the circumcircle of $T_1$ is equal to the radius of the incircle of $T_2$ divided by the radius of the circumcircle of $T_2$. Determine the largest possible value of $b$. Express your answer as a fraction in simplest form.

Solution 1

We can apply the trigonometry theorem $r=4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$

Let $C$ be the apex angle and $A$, $B$ be the base angles of an isosceles traingle.

$\frac{r}{R} = 4\sin\frac{C}{2}\sin^2\frac{B}{2}$

Because $\cos2\theta = 2\cos^2\theta-1 = 1-2\sin^2\theta$, $\sin^2\theta = \frac{1-\cos2\theta}{2}$

$\frac{r}{R} = 4\sin\frac{C}{2}\cdot\frac{1-\cos B}{2}$

The corresponding $\frac{r}{R}$'s for $T_1$ and $T_2$ are:

$T_1: 4\cdot\frac{4}{11}\cdot\frac{1-\frac{4}{11}}{2} = 2\cdot\frac{4}{11}\cdot\frac{7}{11}$

$T_2: 4\cdot\frac{m}{11\\1}\cdot\frac{1-\frac{m}{1}}{2} = 2\cdot m(1-m)$

Therefore $m(1-m) = \frac{4}{11}\cdot\frac{7}{11}$, $m=\frac{7}{11}$.

$b = 2\cdot m = \boxed{\frac{14}{11}}$

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