Difference between revisions of "2022 AIME I Problems/Problem 3"
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MRENTHUSIASM (talk | contribs) (→Solution 1: Made the diagram consistent.) |
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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: | 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> | <asy> | ||
− | + | /* Made by MRENTHUSIASM */ | |
− | pair A = (-250, | + | size(300); |
− | + | pair A, B, C, D, A1, B1, C1, D1, P, Q, P1, Q1; | |
− | + | 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); | |
− | + | P1 = intersectionpoint(A--D,P--(-300)*(Q-P)+P); | |
− | + | Q1 = intersectionpoint(B--C,Q--300*(Q-P)+Q); | |
− | + | draw(anglemark(P,A,B,1000),red); | |
− | + | draw(anglemark(D,A,P,1000),red); | |
− | draw( | + | 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)); | ||
+ | dot("$P'$",P1,1.5*W,linewidth(4)); | ||
+ | dot("$Q'$",Q1,1.5*E,linewidth(4)); | ||
+ | draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); | ||
+ | draw(P--P1^^Q--Q1,dashed); | ||
</asy> | </asy> | ||
− | |||
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>. By symmetry, <math>P'DP</math> is also isosceles, and thus <math>P'A = \frac{AD}{2}</math>. Similarly, the same thing is happening on the right side of the trapezoid, and thus <math>P'Q'</math> is the midline of the trapezoid. Then, <math>PQ = P'Q' - (P'P + Q'Q)</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>. By symmetry, <math>P'DP</math> is also isosceles, and thus <math>P'A = \frac{AD}{2}</math>. Similarly, the same thing is happening on the right side of the trapezoid, and thus <math>P'Q'</math> is the midline of the trapezoid. Then, <math>PQ = P'Q' - (P'P + Q'Q)</math>. | ||
Revision as of 18:19, 18 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
~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
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
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.