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

Revision as of 20:17, 18 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

$[asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot("$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); label("$500$",midpoint(A--B),1.25N); label("$650$",midpoint(C--D),1.25S); label("$333$",midpoint(A--D),1.25W); label("$333$",midpoint(B--C),1.25E); [/asy]$ ~MRENTHUSIASM ~ihatemath123

Solution 1

Extend line $PQ$ to meet $AD$ at $P'$ and $BC$ at $Q'$ . The diagram looks like this: $[asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, P1, Q1; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); P1 = intersectionpoint(A--D,P--(-300)*(Q-P)+P); Q1 = intersectionpoint(B--C,Q--300*(Q-P)+Q); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot("$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'$",Q1,1.5*E,linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); draw(P--P1^^Q--Q1,dashed); [/asy]$ 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

We have the following diagram: $[asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); X = intersectionpoint(B--5*(Q-B)+B,C--D); Y = intersectionpoint(C--5*(Q-C)+C,A--B); Z = intersectionpoint(D--5*(P-D)+D,A--B); W = intersectionpoint(A--5*(P-A)+A,C--D); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot("$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$",P,1.5*(-1,0),linewidth(4)); dot("$Q$",Q,1.5*E,linewidth(4)); dot("$X$",X,1.5*dir(-105),linewidth(4)); dot("$Y$",Y,1.5*N,linewidth(4)); dot("$Z$",Z,1.5*N,linewidth(4)); dot("$W$",W,1.5*dir(-75),linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); draw(P--Z^^P--W^^Q--X^^Q--Y,dashed); [/asy]$ 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

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 $X$ and $Y$ be the feet of the altitudes from $P$ and $Q$ , respectively, to $AB$ , and let $Z$ and $W$ be the feet of the altitudes from $P$ and $Q$ , respectively, to $CD$ . Side $AB$ is parallel to side $CD$ , so $XYWZ$ is a rectangle with width $PQ$ . Furthermore, because $CD - AB = 650-500 = 150$ and trapezoid $ABCD$ is isosceles, $WC - YB = ZD - XA = 75$ .

Also because $ABCD$ is isosceles, $\angle ABC + \angle BCD$ is half the total sum of angles in $ABCD$ , or $180^{\circ}$ . Since $BQ$ and $CQ$ bisect $\angle ABC$ and $\angle BCD$ , respectively, we have $\angle QBC + \angle QCB = 90^{\circ}$ , so $\angle BQC = 90^{\circ}$ .

Letting $BQ = 333k$ , applying Pythagoras to $\triangle BQC$ yields $QC = 333\sqrt{1-k^2}$ . We then proceed using similar triangles: $\angle BYQ = \angle BQC = 90^{\circ}$ and $\angle YBQ = \angle QBC$ , so by AA similarity $YB = 333k^2$ . Likewise, $\angle CWQ = \angle BQC = 90^{\circ}$ and $\angle WCQ = \angle QCB$ , so by AA similarity $WC = 333(1 - k^2)$ . Thus $WC + YB = 333$ .

Adding our two equations for $WC$ and $YB$ gives $2WC = 75 + 333 = 408$ . Therefore, $PQ = ZW = CD - 2WC = 650 - 408 = \boxed{242}$ .

~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)

2022 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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