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Difference between revisions of "Mock AIME 5 2005-2006 Problems/Problem 14"

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== Solution ==
 
== Solution ==
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Notice that <math>X</math> is the reflection of <math>H</math> through the midpoint of <math>BC</math>. So by reflecting the orthocenter lemma we know that <math>AX</math> is a diametre of <math>(ABC)</math>. [<math>(ABC)</math> means circumcircle of <math>\triangle{ABC}</math>. Simillarly <math>BY</math> and <math>CZ</math> also diametre of <math>(ABC)</math>]. So we need to find <math>6R</math> where <math>R</math> is the  radius of <math>(ABC)</math>
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Notice that <math>X</math> is the reflection of <math>H</math> through the midpoint of <math>BC</math>. So by reflecting the orthocenter lemma we know that <math>AX</math> is a diametre of <math>(ABC)</math>. [<math>(ABC)</math> means circumcircle of <math>\triangle{ABC}</math>]. Simillarly <math>BY</math> and <math>CZ</math> also diametre of <math>(ABC)</math>. So we need to find <math>6R</math> where <math>R</math> is the  radius of <math>(ABC)</math>
  
 
Now by cosine rule we get,
 
Now by cosine rule we get,

Revision as of 09:04, 9 August 2022

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

Solution

See also

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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