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Difference between revisions of "2018 MPFG Problem 17"

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==Solution 1==
 
==Solution 1==
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[[File:Mpfg201817.png|600px|center]]
  
 
<math>Y</math>,<math>C</math> and <math>Z</math> is collinear.
 
<math>Y</math>,<math>C</math> and <math>Z</math> is collinear.
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<math>\angle ZCX = 15^{\circ} + 75^{\circ} = 90^{\circ}</math>
 
<math>\angle ZCX = 15^{\circ} + 75^{\circ} = 90^{\circ}</math>
  
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<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>
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<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>
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~cassphe

Latest revision as of 12:30, 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

Mpfg201817.png

$Y$,$C$ 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

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