Difference between revisions of "Mock AIME 5 2005-2006 Problems/Problem 12"

Latest revision as of 15:24, 24 August 2022

Problem

Let $ABC$ be a triangle with $AB = 13$ , $BC = 14$ , and $AC = 15$ . Let $D$ be the foot of the altitude from $A$ to $BC$ and $E$ be the point on $BC$ between $D$ and $C$ such that $BD = CE$ . Extend $AE$ to meet the circumcircle of $ABC$ at $F$ . If the area of triangle $FAC$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers, find $m+n$ .

Solution

Let area of $\triangle XYZ$ be denoted by $[XYZ]$ .

By Heron's theorem , We get

$[ABC]=84$

$Or,\frac{AD.BC}{2}=84$

$Or,AD=12$

By pythagoras theorem on $\triangle ABD$ , We get $BD=5=CE$ So, $DE=4$ .

Again applying pythagoras theorem on $\triangle AED$ , We get, $AE=\sqrt{160}$

As $\triangle ABC$ and $\triangle AEC$ share same height ,

So, $\frac{[AEC]}{[ABC]}=\frac{EC}{BC}=\frac{5}{14}$

Thus $[AEC]=30$

So, $[ABE]=84-30=54$

Now, $\angle AEB = \angle FEC$ [Vertical angle]

$\angle BAF = \angle BCF$ [shares same chord BF]

$Or, \angle BAE = \angle ECF$

So, $\triangle ABE$ is simillar to $\triangle CFE$

Now, $\frac{[CEF]}{[AEB]}=\frac{CE^2}{AE^2}=\frac{25}{160}=\frac{5}{32}$

So, $[CEF]=\frac{135}{16}$

Now, $[ACF]=[CEF]+[AEC]=\frac{135}{16}+30=\frac{615}{16}$

Thus required answer= $615+16=\fbox{631}$

~by NOOBMASTER_M

@@ Line 1: / Line 1: @@
 == Problem ==
+Let <math>ABC</math> be a triangle with <math>AB = 13</math>, <math>BC = 14</math>, and <math>AC = 15</math>. Let <math>D</math> be the foot of the altitude from <math>A</math> to <math>BC</math> and <math>E</math> be the point on <math>BC</math> between <math>D</math> and <math>C</math> such that <math>BD = CE</math>. Extend <math>AE</math> to meet the circumcircle of <math>ABC</math> at <math>F</math>. If the area of triangle <math>FAC</math> is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>m+n</math>.
+== Solution ==
+Let area of <math>\triangle XYZ</math>  be denoted by <math>[XYZ]</math>.
+By Heron's theorem , We get
+<math>[ABC]=84</math>
+<math>Or,\frac{AD.BC}{2}=84</math>
+<math>Or,AD=12</math>
+By pythagoras theorem on <math>\triangle ABD</math> , We get <math>BD=5=CE</math>
+So, <math>DE=4</math>.
+Again applying pythagoras theorem on  <math>\triangle AED</math> , We get, <math>AE=\sqrt{160}</math>
+As  <math>\triangle ABC</math> and  <math>\triangle AEC</math> share same height ,
+So, <math>\frac{[AEC]}{[ABC]}=\frac{EC}{BC}=\frac{5}{14}</math>
+Thus <math>[AEC]=30</math>
+So, <math>[ABE]=84-30=54</math>
+Now, <math>\angle AEB = \angle FEC</math> [Vertical angle]
+<math>\angle BAF = \angle BCF</math> [shares same chord BF]
+<math>Or, \angle BAE = \angle ECF</math>
+So, <math>\triangle ABE</math> is simillar to <math>\triangle CFE</math>
+Now, <math>\frac{[CEF]}{[AEB]}=\frac{CE^2}{AE^2}=\frac{25}{160}=\frac{5}{32}</math>
+So, <math>[CEF]=\frac{135}{16}</math>
+Now, <math>[ACF]=[CEF]+[AEC]=\frac{135}{16}+30=\frac{615}{16}</math>
+Thus required answer=<math>615+16=\fbox{631}</math>
+~by NOOBMASTER_M
 == Solution ==

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

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Difference between revisions of "Mock AIME 5 2005-2006 Problems/Problem 12"

Latest revision as of 15:24, 24 August 2022

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Problem

Solution

Solution

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