2018 MPFG Problem 10: Difference between revisions

Revision as of 05:55, 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$ (Error compiling LaTeX. Unknown error_msg)\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}$

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 ==Problem ==
 Let <imath>T_1</imath> be an isosceles triangle with sides of length <imath>8</imath>, <imath>11</imath>, and <imath>11</imath>. Let <imath>T_2</imath> be an isosceles triangle with sides of length <imath>b</imath>, <imath>1</imath>, and <imath>1</imath>. Suppose that the radius of the incircle of <imath>T_1</imath> divided by the radius of the circumcircle of <imath>T_1</imath> is equal to the radius of the incircle of <imath>T_2</imath> divided by the radius of the circumcircle of <imath>T_2</imath>. Determine the largest possible value of <imath>b</imath>. Express your answer as a fraction in simplest form.
+==Solution 1==
+We can apply the trigonometry theorem <imath>r=4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}
+Let C
+</imath>\frac{r}{R} = 4\sin\frac{C}{2}\sin^2\frac{B}{2}<imath>
+Because </imath>\cos2\theta = 2\cos^2\theta-1 = 1-2\sin^2\theta<imath>, </imath>\sin^2\theta = \frac{1-\cos2\theta}{2}$

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2018 MPFG Problem 10: Difference between revisions

Revision as of 05:55, 8 November 2025

Problem

Solution 1