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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> 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>.
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Since <math>P'P = P'A = \frac{AD}{2}, Q'Q = Q'B = \frac{BC}{2}</math> and <math>AD = BC = 333</math>, we have <math>P'P + Q'Q = \frac{333}{2} + \frac{333}{2} = 333</math>. The length of the midline of a trapezoid is the average of their bases, so <math>P'Q' = \frac{500+650}{2} = 575</math>. Finally, <math>PQ = 575 - 333 = \boxed{242}</math>
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~KingRavi
  
 
== Solution 2==
 
== Solution 2==

Revision as of 19:31, 17 February 2022

Problem

In isosceles trapezoid $ABCD$, parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$, respectively, and $AD=BC=333$. The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$, and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$. Find $PQ$.

Diagram

[asy] unitsize(0.016cm); pair A = (-250,324.4); pair B = (250, 324.4); pair C = (325, 0); pair D = (-325, 0); draw(A--B--C--D--cycle); pair W = (8,0); pair X = (-8, 0); pair Y = (-83,324.4); pair Z = (83,324.4); pair P = (-121, 162.2); pair Q = (121, 162.2); dot(P); dot(Q); draw(A--W, dashed); draw(B--X, dashed); draw(C--Y, dashed); draw(D--Z, dashed); label("$A$", A, N); label("$B$", B, N); label("$Y$", Y, N); label("$Z$", Z, N); label("$C$", C, S); label("$D$", D, S); label("$W$", W, SE); label("$X$", X, SW); label("$P$", P, N); label("$Q$", Q, N); [/asy]

Solution 1

Extend line $PQ$ to meet $AD$ at $P'$ and $BC$ at $Q'$. The diagram looks like this:

[asy] unitsize(0.016cm); pair A = (-250,324.4); pair B = (250, 324.4); pair C = (325, 0); pair D = (-325, 0); draw(A--B--C--D--cycle); pair P1 = (-121, 162.2); pair P2 = (-287.5,162.2); pair Q1 = (121, 162.2); pair Q2 = (287.5,162.2); dot(P1); dot(Q1); dot(P2); dot(Q2); draw(P2--Q2); draw(A--P1, dashed); draw(D--P1, dashed); draw(B--Q1, dashed); draw(C--Q1, dashed); label("$A$", A, N); label("$B$", B, N); label("$C$", C, S); label("$D$", D, S); label("$P$", P1, N); label("$Q$", Q1, N); label("$P'$", P2, W); label("$Q'$", Q2, E); [/asy]

Because the trapezoid is isosceles, by symmetry $PQ$ is parallel to $AB$ and $BC$. Therefore, $\angle PAB \cong \angle APP'$ by interior angles and $\angle PAB \cong \angle PAD$ by the problem statement. Thus, $\triangle P'AP$ is isosceles with $P'P = P'A$. By symmetry, $P'DP$ is also isosceles, and thus $P'A = \frac{AD}{2}$. Similarly, the same thing is happening on the right side of the trapezoid, and thus $P'Q'$ is the midline of the trapezoid. Then, $PQ = P'Q' - (P'P + Q'Q)$.

Since $P'P = P'A = \frac{AD}{2}, Q'Q = Q'B = \frac{BC}{2}$ and $AD = BC = 333$, we have $P'P + Q'Q = \frac{333}{2} + \frac{333}{2} = 333$. The length of the midline of a trapezoid is the average of their bases, so $P'Q' = \frac{500+650}{2} = 575$. Finally, $PQ = 575 - 333 = \boxed{242}$

~KingRavi

Solution 2

Extend lines $AP$ and $BQ$ to meet line $DC$ at points $W$ and $X$, respectively, and extend lines $DP$ and $CQ$ to meet $AB$ at points $Z$ and $Y$, respectively.

Claim: quadrilaterals $AZWD$ and $BYXD$ are rhombuses.

Proof: Since $\angle DAB + \angle ADC = 180^{\circ}$, $\angle ADP + \angle PAD = 90^{\circ}$. Therefore, triangles $APD$, $APZ$, $DPW$ and $PZW$ are all right triangles. By SAA congruence, the first three triangles are congruent; by SAS congruence, $\triangle PZW$ is congruent to the other three. Therefore, $AD = DW = WZ = AZ$, so $AZWD$ is a rhombus. By symmetry, $BYXC$ is also a rhombus.

Extend line $PQ$ to meet $\overline{AD}$ and $\overline{BC}$ at $R$ and $S$, respectively. Because of rhombus properties, $RP = QS = \frac{333}{2}$. Also, by rhombus properties, $R$ and $S$ are the midpoints of segments $AD$ and $BC$, respectively; therefore, by trapezoid properties, $RS = \frac{AB + CD}{2} = 575$. Finally, $PQ = RS - RP - QS = \boxed{242}$.

~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

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