Difference between revisions of "2025 AIME I Problems"
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==Problem 2== | ==Problem 2== | ||
| − | [[2025 AIME I Problems/Problem 2|Solution]] | + | In <math>\triangle ABC</math> points <math>D</math> and <math>E</math> lie on <math>\overline{AB}</math> so that <math>AD < AE < AB</math>, while points <math>F</math> and <math>G</math> lie on <math>\overline{AC}</math> so that <math>AF < AG < AC</math>. Suppose <math>AD = 4</math>, <math>DE = 16</math>, <math>EB = 8</math>, <math>AF = 13</math>, <math>FG = 52</math>, and <math>GC = 26</math>. Let <math>M</math> be the reflection of <math>D</math> through <math>F</math>, and let <math>N</math> be the reflection of <math>G</math> through <math>E</math>. The area of quadrilateral <math>DEGF</math> is <math>288</math>. Find the area of heptagon <math>AFNBCEM</math>, as shown in the figure below. |
| + | |||
| + | <asy> | ||
| + | unitsize(14); | ||
| + | pair A = (0, 9), B = (-6, 0), C = (12, 0), D = (5A + 2B)/7, E = (2A + 5B)/7, F = (5A + 2C)/7, G = (2A + 5C)/7, M = 2F - D, N = 2E - G; | ||
| + | filldraw(A--F--N--B--C--E--M--cycle, lightgray); | ||
| + | draw(A--B--C--cycle); | ||
| + | draw(D--M); | ||
| + | draw(N--G); | ||
| + | dot(A); | ||
| + | dot(B); | ||
| + | dot(C); | ||
| + | dot(D); | ||
| + | dot(E); | ||
| + | dot(F); | ||
| + | dot(G); | ||
| + | dot(M); | ||
| + | dot(N); | ||
| + | label("$A$", A, dir(90)); | ||
| + | label("$B$", B, dir(225)); | ||
| + | label("$C$", C, dir(315)); | ||
| + | label("$D$", D, dir(135)); | ||
| + | label("$E$", E, dir(135)); | ||
| + | label("$F$", F, dir(45)); | ||
| + | label("$G$", G, dir(45)); | ||
| + | label("$M$", M, dir(45)); | ||
| + | label("$N$", N, dir(135)); | ||
| + | </asy> | ||
| + | |||
| + | [[2025 AIME I Problems/Problem 2|Solution]] | ||
==Problem 3== | ==Problem 3== | ||
Revision as of 17:07, 13 February 2025
| 2025 AIME I (Answer Key) | AoPS Contest Collections • PDF | ||
|
Instructions
| ||
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
to 
, inclusive. Your score will be the number of correct answers; i.e., there is neither partial credit nor a penalty for wrong answers.Contents
Problem 1
Find the sum of all integer bases 
for which 
is a divisor of 
.
Problem 2
In 
points 
and 
lie on 
so that 
, while points 
and 
lie on 
so that 
. Suppose 
, 
, 
, 
, 
, and 
. Let 
be the reflection of through , and let 
be the reflection of through . The area of quadrilateral 
is 
. Find the area of heptagon 
, as shown in the figure below.
![[asy] unitsize(14); pair A = (0, 9), B = (-6, 0), C = (12, 0), D = (5A + 2B)/7, E = (2A + 5B)/7, F = (5A + 2C)/7, G = (2A + 5C)/7, M = 2F - D, N = 2E - G; filldraw(A--F--N--B--C--E--M--cycle, lightgray); draw(A--B--C--cycle); draw(D--M); draw(N--G); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(M); dot(N); label("$A$", A, dir(90)); label("$B$", B, dir(225)); label("$C$", C, dir(315)); label("$D$", D, dir(135)); label("$E$", E, dir(135)); label("$F$", F, dir(45)); label("$G$", G, dir(45)); label("$M$", M, dir(45)); label("$N$", N, dir(135)); [/asy]](http://latex.artofproblemsolving.com/6/4/b/64b0d84a95e0388e37622b54a251d836088fd97c.png)
Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12
Problem 13
Problem 14
Problem 15
See also
| 2025 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by 2024 AIME II |
Followed by 2025 AIME II | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
- American Invitational Mathematics Examination
- AIME Problems and Solutions
- Mathematics competition resources
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
