2022 AIME I Problems/Problem 3
Contents
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
![[asy] 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 W = (8,0); pair X = (-8, 0); pair Y = (-83,324.4); pair Z = (83,324.4); pair P = (-121, 162.2); pair Q = (121, 162.2); dot(P); dot(Q); draw(A--W, dashed); draw(B--X, dashed); draw(C--Y, dashed); draw(D--Z, dashed); label("$A$", A, N); label("$B$", B, N); label("$Y$", Y, N); label("$Z$", Z, N); label("$C$", C, S); label("$D$", D, S); label("$W$", W, SE); label("$X$", X, SW); label("$P$", P, N); label("$Q$", Q, N); [/asy]](http://latex.artofproblemsolving.com/8/8/1/8817bea003e7bb26ea4371cdf8ba8bbf940f354e.png)
Solution
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
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 | ||
