# Difference between revisions of "1995 AIME Problems/Problem 9"

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== Solution == | == Solution == | ||

− | Let <math>\angle CAM=x</math>, <math>\ | + | Let <math>\angle CAM=x</math>, <math>\angle CDM=3x</math>. Then, <math>(tan 3x)/(tan x)=(CM/1)/(CM/11)=11</math>. Expanding tan 3x using the angle sum formula gives <math>tan 3x=(3tan x-tan^3x)/(1-3tan^2x)</math>. Thus, <math>(3-tan^2x)/(1-3tan^2x)=11</math>. Solving, we get <math>tan x=1/2</math>. Hence, <math>CM=11/2</math> and AC=<math>11\sqrt{5}/2</math> by Pythag. The total perimeter is double the sum of these, which is <math>\sqrt{605}+11</math>. The answer is then <math>616</math>. |

== See also == | == See also == |

## Revision as of 20:15, 17 June 2008

## Problem

Triangle is isosceles, with and altitude Suppose that there is a point on with and Then the perimeter of may be written in the form where and are integers. Find

## Solution

Let , . Then, . Expanding tan 3x using the angle sum formula gives . Thus, . Solving, we get . Hence, and AC= by Pythag. The total perimeter is double the sum of these, which is . The answer is then .