Difference between revisions of "2018 AIME I Problems/Problem 4"

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==Problem 4==
 
==Problem 4==
 
In <math>\triangle ABC, AB = AC = 10</math> and <math>BC = 12</math>. Point <math>D</math> lies strictly between <math>A</math> and <math>B</math> on <math>\overline{AB}</math> and point <math>E</math> lies strictly between <math>A</math> and <math>C</math> on <math>\overline{AC}</math>) so that <math>AD = DE = EC</math>. Then <math>AD</math> can be expressed in the form <math>\dfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.
 
In <math>\triangle ABC, AB = AC = 10</math> and <math>BC = 12</math>. Point <math>D</math> lies strictly between <math>A</math> and <math>B</math> on <math>\overline{AB}</math> and point <math>E</math> lies strictly between <math>A</math> and <math>C</math> on <math>\overline{AC}</math>) so that <math>AD = DE = EC</math>. Then <math>AD</math> can be expressed in the form <math>\dfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.
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==Solution 1==

Revision as of 18:40, 7 March 2018

Problem 4

In $\triangle ABC, AB = AC = 10$ and $BC = 12$. Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$) so that $AD = DE = EC$. Then $AD$ can be expressed in the form $\dfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

Solution 1