2022 AIME I Problems/Problem 3
Contents
[hide]Problem
In isosceles trapezoid , parallel bases and have lengths and , respectively, and . The angle bisectors of and meet at , and the angle bisectors of and meet at . Find .
Diagram
Solution 1
Extend line to meet at and at . The diagram looks like this:
unitsize(0.016cm); pair A = (-250,324.4); pair B = (250, 324.4); pair C = (325, 0); pair D = (-325, 0); draw(A--B--C--D--cycle); pair P = (-121, 162.2); pair P' =(-412.5,162.2); pair Q = (121, 162.2); pair Q' = (412.5,162.5); dot(P); dot(Q); draw(P'--Q'); label("$A$", A, N); label("$B$", B, N); label("$C$", C, S); label("$D$", D, S); label("$P$", P, N); label("$Q$", Q, N); label("$P'$", P', W); label($"Q'$", Q', E); (Error making remote request. Unknown error_msg)
Solution 2
Extend lines and to meet line at points and , respectively, and extend lines and to meet at points and , respectively.
Claim: quadrilaterals and are rhombuses.
Proof: Since , . Therefore, triangles , , and are all right triangles. By SAA congruence, the first three triangles are congruent; by SAS congruence, is congruent to the other three. Therefore, , so is a rhombus. By symmetry, is also a rhombus.
Extend line to meet and at and , respectively. Because of rhombus properties, . Also, by rhombus properties, and are the midpoints of segments and , respectively; therefore, by trapezoid properties, . Finally, .
~ihatemath123
Video Solution (Mathematical Dexterity)
https://www.youtube.com/watch?v=fNAvxXnvAxs
See Also
2022 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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