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Difference between revisions of "2022 AIME I Problems/Problem 3"

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== Problem ==
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In isosceles trapezoid <math>ABCD</math>, parallel bases <math>\overline{AB}</math> and <math>\overline{CD}</math> have lengths <math>500</math> and <math>650</math>, respectively, and <math>AD=BC=333</math>. The angle bisectors of <math>\angle{A}</math> and <math>\angle{D}</math> meet at <math>P</math>, and the angle bisectors of <math>\angle{B}</math> and <math>\angle{C}</math> meet at <math>Q</math>. Find <math>PQ</math>.
  
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== Diagram ==
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<asy>
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unitsize(0.016cm);
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pair A = (-250,324.4);
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pair B = (250, 324.4);
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pair C = (325, 0);
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pair D = (-325, 0);
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draw(A--B--C--D--cycle);
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pair W = (8,0);
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pair X = (-8, 0);
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pair Y = (-83,324.4);
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pair Z = (83,324.4);
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pair P = (-121, 162.2);
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pair Q = (121, 162.2);
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dot(P);
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dot(Q);
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draw(A--W, dashed);
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draw(B--X, dashed);
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draw(C--Y, dashed);
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draw(D--Z, dashed);
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label("$A$", A, N);
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label("$B$", B, N);
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label("$Y$", Y, N);
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label("$Z$", Z, N);
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label("$C$", C, S);
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label("$D$", D, S);
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label("$W$", W, SE);
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label("$X$", X, SW);
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label("$P$", P, N);
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label("$Q$", Q, N);
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</asy>
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Extend lines <math>AP</math> and <math>BQ</math> to meet line <math>DC</math> at points <math>W</math> and <math>X</math>, respectively, and extend lines <math>DP</math> and <math>CQ</math> to meet <math>AB</math> at points <math>Z</math> and <math>Y</math>, respectively.
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Claim: quadrilaterals <math>AZWD</math> and <math>BYXD</math> are rhombuses.
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Proof: Since <math>\angle DAB + \angle ADC = 180^{\circ}</math>, <math>\angle ADP + \angle PAD = 90^{\circ}</math>. Therefore, triangles <math>APD</math>, <math>APZ</math>, <math>DPW</math> and <math>PZW</math> are all right triangles. By SAA congruence, the first three triangles are congruent; by SAS congruence, <math>\triangle PZW</math> is congruent to the other three. Therefore, <math>AD = DW = WZ = AZ</math>, so <math>AZWD</math> is a rhombus. By symmetry, <math>BYXC</math> is also a rhombus.
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Extend line <math>PQ</math> to meet <math>\overline{AD}</math> and <math>\overline{BC}</math> at <math>R</math> and <math>S</math>, respectively. Because of rhombus properties, <math>RP = QS = \frac{333}{2}</math>. Also, by rhombus properties, <math>R</math> and <math>S</math> are the midpoints of segments <math>AD</math> and <math>BC</math>, respectively; therefore, by trapezoid properties, <math>RS = \frac{AB + CD}{2} = 575</math>. Finally, <math>PQ = RS - RP - QS = \boxed{242}</math>.
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~ihatemath123

Revision as of 16:07, 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]

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

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