Talk:2021 AIME I Problems/Problem 9: Difference between revisions
MRENTHUSIASM (talk | contribs) Created page with "Based on Solution 1: Note: Instead of solving the system of equations (1)(2) which can be time consuming, by noting that <math>\triangle ACF \sim \triangle ABG</math> by AA,..." |
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However, I don't think <math>\triangle ACF \sim \triangle ABG</math> is convincing. How do you find that by AA? | However, I don't think <math>\triangle ACF \sim \triangle ABG</math> is convincing. How do you find that by AA? | ||
~MRENTHUSIASM | |||
Revision as of 03:06, 24 January 2024
Based on Solution 1:
Note: Instead of solving the system of equations (1)(2) which can be time consuming, by noting that 
by AA, we could find out 
, which gives 
. We also know that 
by Pythagorean Theorem on 
. 
. Then using Pythagorean Theorem on 
we obtain:
![\[AC^2 = (EB+BC)^2 + AE^2\]](http://latex.artofproblemsolving.com/b/f/d/bfdf330ba1fdc7cfed58c82d6cae00b338f44bda.png)
substituting, we get:
![\[\frac{81}{25}x^2 = (\sqrt{x^2 -225}+\frac{6}{5}x)^2+225 \iff x = 3\sqrt{x^2 - 15^2}\]](http://latex.artofproblemsolving.com/5/4/3/543b1f43bcb9024eb36908d428110f4a23740f88.png)
Finally, we solve to obtain 
.
~Chupdogs
However, I don't think is convincing. How do you find that by AA?
~MRENTHUSIASM
