Difference between revisions of "2022 AIME I Problems/Problem 3"
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+ | == Solution 3 == | ||
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+ | Let <math>X</math> and <math>Y</math> be the feet of the altitudes from <math>P</math> and <math>Q</math>, respectively, to <math>AB</math>, and let <math>Z</math> and <math>W</math> be the feet of the altitudes from <math>P</math> and <math>Q</math>, respectively, to <math>CD</math>. Side <math>AB</math> is parallel to side <math>CD</math>, so <math>XYWZ</math> is a rectangle with width <math>PQ</math>. Furthermore, because <math>CD - AB = 650-500 = 150</math> and trapezoid <math>ABCD</math> is isosceles, <math>WC - YB = ZD - XA = 75</math>. | ||
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+ | Also because <math>ABCD</math> is isosceles, <math>\angle ABC + \angle BCD</math> is half the total sum of angles in <math>ABCD</math>, or <math>180^{\circ}</math>. Since <math>BQ</math> and <math>CQ</math> bisect <math>\angle ABC</math> and <math>\angle BCD</math>, respectively, we have <math>\angle QBC + \angle QCB = 90^{\circ}</math>, so <math>\angle BQC = 90^{\circ}</math>. | ||
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+ | Letting <math>BQ = 333k</math>, applying Pythagoras to <math>\triangle BQC</math> yields <math>QC = 333\sqrt{1-k^2}</math>. We then proceed using similar triangles: <math>\angle BYQ = \angle BQC = 90^{\circ}</math> and <math>\angle YBQ = \angle QBC</math>, so by AA similarity <math>YB = 333k^2</math>. Likewise, <math>\angle CWQ = \angle BQC = 90^{\circ}</math> and <math>\angle WCQ = \angle QCB</math>, so by AA similarity <math>WC = 333(1 - k^2)</math>. Thus <math>WC + YB = 333</math>. | ||
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+ | Adding our two equations for <math>WC</math> and <math>YB</math> gives <math>2WC = 75 + 333 = 408</math>. Therefore, <math>PQ = ZW = CD - 2WC = 650 - 408 = \boxed{242}</math>. | ||
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+ | ~Orange_Quail_9 | ||
==Video Solution (Mathematical Dexterity)== | ==Video Solution (Mathematical Dexterity)== |
Revision as of 18:32, 18 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
~MRENTHUSIASM ~ihatemath123
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 . By symmetry, is also isosceles, and thus . Similarly, the same thing is happening on the right side of the trapezoid, and thus is the midline of the trapezoid. Then, .
Since and , we have . The length of the midline of a trapezoid is the average of their bases, so . Finally,
~KingRavi
Solution 2
We have the following diagram: 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
Solution 3
Let and be the feet of the altitudes from and , respectively, to , and let and be the feet of the altitudes from and , respectively, to . Side is parallel to side , so is a rectangle with width . Furthermore, because and trapezoid is isosceles, .
Also because is isosceles, is half the total sum of angles in , or . Since and bisect and , respectively, we have , so .
Letting , applying Pythagoras to yields . We then proceed using similar triangles: and , so by AA similarity . Likewise, and , so by AA similarity . Thus .
Adding our two equations for and gives . Therefore, .
~Orange_Quail_9
Video Solution (Mathematical Dexterity)
https://www.youtube.com/watch?v=fNAvxXnvAxs
Video Solution
https://www.youtube.com/watch?v=h_LOT-rwt08
~Steven Chen (www.professorchenedu.com)
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.