2018 MPFG Problem 17: Difference between revisions

Revision as of 11:30, 29 August 2025

Problem

Let $ABC$ be a triangle with $AB = 5$ , $BC = 4$ , and $CA = 3$ . On each side of $ABC$ , externally erect a semicircle whose diameter is the corresponding side. Let $X$ be on the semicircular arc erected on side $BC$ such that $\angle CBX$ has measure $15^{\circ}$ . Let $Y$ be on the semicircular arc erected on side $CA$ such that $\angle ACY$ has measure $15^{\circ}$ . Similarly, let $Z$ be on the semicircular arc erected on side $AB$ such that $\angle BAZ$ has measure $15^{\circ}$ . What is the area of triangle $\Delta XYZ$ ? Express your answer as a fraction in simplest form.

Solution 1

$Y$ , $C$ and $Z$ is collinear.

Because $\angle ACB = \angle AZB = 90^{\circ}$ , $ACBZ$ is concyclic. $\angle ZCB = \angle ZAB = 15^{\circ}$

$\angle ZCX = 15^{\circ} + 75^{\circ} = 90^{\circ}$

$S_{\Delta XYZ} = S_{ACBZ} = S_{\Delta ACB} + S_{\Delta AZB} = 6 + \frac{1}{2} \cdot 5^2 \cdot sin15^{\circ}cos15^{\circ} = 6 + \frac{1}{2} \cdot 5^2 \cdot \frac{1}{2}sin30^{\circ} = 6+\frac{25}{8} = \boxed{\frac{73}{8}}$

~cassphe

@@ Line 12: / Line 12: @@
 <math>\angle ZCX = 15^{\circ} + 75^{\circ} = 90^{\circ}</math>
-<math>S_{\Delta XYZ} = S_{ACBZ} = S_{\Delta ACB} + S_{\Delta AZB} = 6 + \frac{1}{2} \cdot 5^2 \cdot sin15^{\circ}cos15^{\circ} = 6 + \frac{1}{2} \cdot 5^2 \cdot \frac{1}{2}sin30^{\circ} = 6+\frac{25}{8} = \frac{73}{8}</math>
+<math>S_{\Delta XYZ} = S_{ACBZ} = S_{\Delta ACB} + S_{\Delta AZB} = 6 + \frac{1}{2} \cdot 5^2 \cdot sin15^{\circ}cos15^{\circ} = 6 + \frac{1}{2} \cdot 5^2 \cdot \frac{1}{2}sin30^{\circ} = 6+\frac{25}{8} = \boxed{\frac{73}{8}}</math>
+~cassphe

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2018 MPFG Problem 17: Difference between revisions

Revision as of 11:30, 29 August 2025

Problem

Solution 1