2022 AIME I Problems/Problem 3: Difference between revisions
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== Solution 1== | == Solution 1 == | ||
Extend line <math>PQ</math> to meet <math>AD</math> at <math>P'</math> and <math>BC</math> at <math>Q'</math>. The diagram looks like this: | |||
<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 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); | |||
</asy> | |||
== Solution 2== | |||
Extend lines <math>AP</math> and <math>BQ</math> to meet line <math>DC</math> at points <math>W</math> and <math>X</math>, respectively, and extend lines <math>DP</math> and <math>CQ</math> to meet <math>AB</math> at points <math>Z</math> and <math>Y</math>, respectively. | Extend lines <math>AP</math> and <math>BQ</math> to meet line <math>DC</math> at points <math>W</math> and <math>X</math>, respectively, and extend lines <math>DP</math> and <math>CQ</math> to meet <math>AB</math> at points <math>Z</math> and <math>Y</math>, respectively. | ||
Revision as of 19:10, 17 February 2022
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 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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
