Difference between revisions of "2022 AIME I Problems/Problem 3"
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(→Solution 1) |
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pair P1 = (-121, 162.2); | pair P1 = (-121, 162.2); | ||
− | pair P2 = (- | + | pair P2 = (-287.5,162.2); |
pair Q1 = (121, 162.2); | pair Q1 = (121, 162.2); | ||
− | pair Q2 = ( | + | pair Q2 = (287.5,162.2); |
dot(P1); | dot(P1); | ||
dot(Q1); | dot(Q1); |
Revision as of 20:15, 17 February 2022
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
Solution 1
Extend line to meet
at
and
at
. The diagram looks like this:
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.