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

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== Problem ==
 
== Problem ==
 +
Let <math>ABC</math> be a triangle such that <math>AB = 68</math>, <math>BC = 100</math>, and <math>CA = 112</math>. Let <math>H</math> be the orthocenter of <math>\triangle ABC</math> (intersection of the altitudes). Let <math>D</math> be the midpoint of <math>BC</math>, <math>E</math> be the midpoint of <math>CA</math>, and <math>F</math> be the midpoint of <math>AB</math>. Points <math>X</math>, <math>Y</math>, and <math>Z</math> are constructed on <math>HD</math>, <math>HE</math>, and <math>HF</math>, respectively, such that <math>D</math> is the midpoint of <math>XH</math>, <math>E</math> is the midpoint of <math>YH</math>, and <math>F</math> is the midpoint of <math>ZH</math>. Find <math>AX+BY+CZ</math>.
  
 
== Solution ==
 
== Solution ==
 +
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,
 +
<math>CosB=\frac{68^2+100^2-112^2}{2.68.100}=\frac{13}{85}</math>
 +
 +
So <math>Sin^2B=1-Cos^2B=\frac{85^2-13^2}{85^2}=\frac{84^2}{85^2}</math>
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 +
<math>Or, SinB=\frac{84}{85}</math>
 +
 +
Now by sine rule we get,
 +
 +
<math>2R=\frac{AC}{SinB}=\frac{112}{\frac{84}{85}}=\frac{340}{3}</math>
 +
 +
So required answer is <math>6R=\fbox{340}</math>
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 +
By NOOBMASTER_M
  
 
== See also ==
 
== See also ==

Latest revision as of 09:06, 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

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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