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Difference between revisions of "Talk:2021 AIME I Problems/Problem 9"
(Undo revision 212537 by MRENTHUSIASM (talk))
(Tag: Undo)
 
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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?
  
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~MRENTHUSIASM
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One right angle is shared and <math>\angle ABD = \angle BDC = \angle ACD</math> by alternate angles theorem (for first part) and symmetry for the second part as its an isosceles trapezoid (<math>BD</math> and <math>AC</math> are both diagonals and the angles it makes with the base edge is the same)

Latest revision as of 10:57, 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 $\triangle ACF \sim \triangle ABG$ by AA, we could find out $\frac{AB}{AG} = \frac{AC}{AF}$, which gives $AC = \frac{9}{5}x$. We also know that $EB = \sqrt{x^2 - 15^2}$ by Pythagorean Theorem on $\triangle ABE$. $BC = AD = \frac{6}{5}x$. Then using Pythagorean Theorem on $\triangle ACE$ we obtain: \[AC^2 = (EB+BC)^2 + AE^2\] substituting, we get: \[\frac{81}{25}x^2 = (\sqrt{x^2 -225}+\frac{6}{5}x)^2+225 \iff x = 3\sqrt{x^2 - 15^2}\] Finally, we solve to obtain $x = \frac{45\sqrt{2}}{4}$.

~Chupdogs

However, I don't think $\triangle ACF \sim \triangle ABG$ is convincing. How do you find that by AA?

One right angle is shared and $\angle ABD = \angle BDC = \angle ACD$ by alternate angles theorem (for first part) and symmetry for the second part as its an isosceles trapezoid ($BD$ and $AC$ are both diagonals and the angles it makes with the base edge is the same)

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