2018 MPFG Problem 10: Difference between revisions

Revision as of 07:24, 8 November 2025

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 $\frac{r}[R}$ (Error compiling LaTeX. Unknown error_msg)'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 $m(1-m) = \frac{4}{11}\cdot\frac{7}{11}$ , $m=\frac{7}{11}$ .

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

@@ Line 12: / Line 12: @@
 Because <imath>\cos2\theta = 2\cos^2\theta-1 = 1-2\sin^2\theta</imath>, <imath>\sin^2\theta = \frac{1-\cos2\theta}{2}</imath>
-<imath>\frac{r}{R} = 4\sin\frac{C}{2}\cdot\frac{1-\cosB}{2}</imath>
+<imath>\frac{r}{R} = 4\sin\frac{C}{2}\cdot\frac{1-\cos B}{2}</imath>
 The corresponding <imath>\frac{r}[R}</imath>'s for <imath>T_1</imath> and <imath>T_2</imath> are:

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2018 MPFG Problem 10: Difference between revisions

Revision as of 07:24, 8 November 2025

Problem

Solution 1