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

Revision as of 19:26, 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 $P'P = P'A$ .

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

@@ Line 78: / Line 78: @@
 </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>
+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>.
 == Solution 2==

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