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2018 MPFG Problem 17: Difference between revisions

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==Problem==
==Problem==
Let <math>ABC</math> be a triangle with <math>AB = 5</math>, <math>BC = 4</math>, and <math>CA = 3</math>. On each side of <math>ABC</math>, externally erect a semicircle whose diameter is the corresponding side. Let <math>X</math> be on the semicircular arc erected on side <math>BC</math> such that <math>\angle CBX</math> has measure <math>15^{\circ}</math>. Let <math>Y</math> be on the semicircular arc erected on side <math>CA</math> such that <math>\angle ACY</math> has measure <math>15^{\circ}</math>. Similarly, let <math>Z</math> be on the semicircular arc erected on side <math>AB</math> such that <math>\angle BAZ</math> has measure <math>15^{\circ}</math>. What is the area of triangle <math>\Delta XYZ</math>? Express your answer as a fraction in simplest form.
Let <math>ABC</math> be a triangle with <math>AB = 5</math>, <math>BC = 4</math>, and <math>CA = 3</math>. On each side of <math>ABC</math>, externally erect a semicircle whose diameter is the corresponding side. Let <math>X</math> be on the semicircular arc erected on side <math>BC</math> such that <math>\angle CBX</math> has measure <math>15^{\circ}</math>. Let <math>Y</math> be on the semicircular arc erected on side <math>CA</math> such that <math>\angle ACY</math> has measure <math>15^{\circ}</math>. Similarly, let <math>Z</math> be on the semicircular arc erected on side <math>AB</math> such that <math>\angle BAZ</math> has measure <math>15^{\circ}</math>. What is the area of triangle <math>\Delta XYZ</math>? Express your answer as a fraction in simplest form.
==Solution 1==
[[File:Mpfg201817.png|600px|center]]
<imath>Y</imath>,<imath>C</imath> and <imath>Z</imath> is collinear.
Because <imath>\angle ACB = \angle AZB = 90^{\circ}</imath>, <imath>ACBZ</imath> is concyclic. <imath>\angle ZCB = \angle ZAB = 15^{\circ}</imath>
<imath>\angle ZCX = 15^{\circ} + 75^{\circ} = 90^{\circ}</imath>
<imath>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}}</imath>
~cassphe

Latest revision as of 09:39, 7 November 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

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