Difference between revisions of "2022 AIME I Problems/Problem 3"
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== Solution 3 == | == Solution 3 == | ||
+ | |||
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(300); | ||
+ | pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; | ||
+ | A = (-250,6*sqrt(731)); | ||
+ | B = (250,6*sqrt(731)); | ||
+ | C = (325,-6*sqrt(731)); | ||
+ | D = (-325,-6*sqrt(731)); | ||
+ | A1 = bisectorpoint(B,A,D); | ||
+ | B1 = bisectorpoint(A,B,C); | ||
+ | C1 = bisectorpoint(B,C,D); | ||
+ | D1 = bisectorpoint(A,D,C); | ||
+ | P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); | ||
+ | Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); | ||
+ | draw(anglemark(P,A,B,1000),red); | ||
+ | draw(anglemark(D,A,P,1000),red); | ||
+ | draw(anglemark(A,B,Q,1000),red); | ||
+ | draw(anglemark(Q,B,C,1000),red); | ||
+ | draw(anglemark(P,D,A,1000),red); | ||
+ | draw(anglemark(C,D,P,1000),red); | ||
+ | draw(anglemark(Q,C,D,1000),red); | ||
+ | draw(anglemark(B,C,Q,1000),red); | ||
+ | add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); | ||
+ | add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); | ||
+ | add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); | ||
+ | add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); | ||
+ | add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); | ||
+ | add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); | ||
+ | add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); | ||
+ | add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); | ||
+ | dot("$A$",A,1.5*dir(A),linewidth(4)); | ||
+ | dot("$B$",B,1.5*dir(B),linewidth(4)); | ||
+ | dot("$C$",C,1.5*dir(C),linewidth(4)); | ||
+ | dot("$D$",D,1.5*dir(D),linewidth(4)); | ||
+ | dot("$P$",P,1.5*NE,linewidth(4)); | ||
+ | dot("$Q$",Q,1.5*NW,linewidth(4)); | ||
+ | draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); | ||
+ | X = (-121,6*sqrt(731)); | ||
+ | Y = (121,6*sqrt(731)); | ||
+ | Z = (-121,-6*sqrt(731)); | ||
+ | W = (121,-6*sqrt(731)); | ||
+ | draw(X--Z); | ||
+ | draw(Y--W); | ||
+ | draw(rightanglemark(A,X,Z,500),red); | ||
+ | draw(rightanglemark(B,Y,W,500),red); | ||
+ | draw(rightanglemark(C,W,Y,500),red); | ||
+ | draw(rightanglemark(D,Z,X,500),red); | ||
+ | dot("$X$",X,1.5*N,linewidth(4)); | ||
+ | dot("$Y$",Y,1.5*N,linewidth(4)); | ||
+ | dot("$Z$",Z,1.5*S,linewidth(4)); | ||
+ | dot("$W$",W,1.5*S,linewidth(4)); | ||
+ | </asy> | ||
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>. | 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>. |
Revision as of 20:17, 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.