Difference between revisions of "2024 AIME I Problems/Problem 14"
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| + | Let <math>ABC</math> be a tringel with circumaradius <math>17</math> such that <math>B</math> is a right angle. Find <math>AC</math>. | ||
| + | ==Solution== | ||
| + | Since the circumradius of a right triangle is half the hypotenuse, the answer is <math>\boxed{34}</math>. I lost the game. | ||
| + | |||
| + | ==See also== | ||
| + | {{AIME box|year=2024|n=I|before=[[2023 AIME I]], [[2023 AIME II|II]]|after=[[2024 AIME II]], [[2025 AIME I]], [[2025 AIME II|II]]}} | ||
| + | {{MAA Notice}} | ||
Revision as of 15:20, 31 January 2024
Let 
be a tringel with circumaradius 
such that 
is a right angle. Find 
.
Solution
Since the circumradius of a right triangle is half the hypotenuse, the answer is 
. I lost the game.
See also
| 2024 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by 2023 AIME I, II |
Followed by 2024 AIME II, 2025 AIME I, II | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
