Difference between revisions of "2022 AIME I Problems/Problem 3"
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− | Because the trapezoid is isosceles, by symmetry <math>PQ</math> is parallel to <math>AB</math> and <math>BC</math>. Therefore, <math>\angle PAB \cong \angle APP'</math> | + | Because the trapezoid is isosceles, by symmetry <math>PQ</math> is parallel to <math>AB</math> and <math>BC</math>. Therefore, <math>\angle PAB \cong \angle APP'</math> by interior angles and <math>\angle PAB \cong \angle PAD</math> by the problem statement. Thus, <math>\triangle P'AP</math> is isosceles with <math>P'P = P'A</math>. |
== Solution 2== | == Solution 2== |
Revision as of 19:26, 17 February 2022
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:
Because the trapezoid is isosceles, by symmetry is parallel to and . Therefore, by interior angles and by the problem statement. Thus, is isosceles with .
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.