Mock AIME 5 2005-2006 Problems/Problem 14

Problem

Let $ABC$ be a triangle such that $AB = 68$ , $BC = 100$ , and $CA = 112$ . Let $H$ be the orthocenter of $\triangle ABC$ (intersection of the altitudes). Let $D$ be the midpoint of $BC$ , $E$ be the midpoint of $CA$ , and $F$ be the midpoint of $AB$ . Points $X$ , $Y$ , and $Z$ are constructed on $HD$ , $HE$ , and $HF$ , respectively, such that $D$ is the midpoint of $XH$ , $E$ is the midpoint of $YH$ , and $F$ is the midpoint of $ZH$ . Find $AX+BY+CZ$ .

Solution

Notice that $X$ is the reflection of $H$ through the midpoint of $BC$ . So by reflecting the orthocenter lemma we know that $AX$ is a diametre of $(ABC)$ . [ $(ABC)$ means circumcircle of $\triangle{ABC}$ . Simillarly $BY$ and $CZ$ also diametre of $(ABC)$ ]. So we need to find $6R$ where $R$ is the radius of $(ABC)$

Now by cosine rule we get, $CosB=\frac{68^2+100^2-112^2}{2.68.100}=\frac{13}{85}$

So $Sin^2B=1-Cos^2B=\frac{85^2-13^2}{85^2}=\frac{84^2}{85^2}$

$Or, SinB=\frac{84}{85}$

Now by sine rule we get,

$2R=\frac{AC}{SinB}=\frac{112}{\frac{84}{85}}=\frac{340}{3}$

So required answer is $6R=\fbox{340}$

By NOOBMASTER_M

Mock AIME 5 2005-2006 (Problems, Source)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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Mock AIME 5 2005-2006 Problems/Problem 14

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Problem

Solution

Solution

See also