2024 AIME I Problems/Problem 5
Solution 1 (need help with diagram)
Suppose . Extend and until they meet at . From Power of a Point, we have . Substituting in these values, we get . Using simple guess and check, we find that so .
Solution 2
We use simple geometry to solve this problem.
We are given that , , , and are concyclic; call the circle that they all pass through circle with center . We know that, given any chord on a circle, the perpendicular bisector to the chord passes through the center; thus, given two chords, taking the intersection of their perpendicular bisectors gives the center. We therefore consider chords and and take the midpoints of and to be and , respectively.
We could draw the circumcircle, but actually it does not matter for our solution; all that matters is that , where is the circumradius.
By the Pythagorean Theorem, . Also, . We know that , and ; ; ; and finally, . Let . We now know that and . Recall that ; thus, . We solve for :
\begin{align*} (x+92)^2+8^2&=25^2+92^2 \\ (x+92)^2&=625+(100-8)^2-8^2 \\ &=625+10000-1600+64-64 \\ &=9025 \\ x+92&=95 \\ x&=3. \\ \end{align*}
The question asks for , which is .
~Technodoggo
Someone please edit this solution. I'm bad at latex
Draw a line from A to G
Draw cyclic quad
Triangle formed by A, G, and intersection between lines AB and GF is similar to triangle DHE
Solving similarity ratios gives DE = 3, so CE = 107 - 3 = 104 ~coolruler
See also
2024 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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