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 $PQ$

Solution: [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^^Y--W,dashed); 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, the answer is $PQ = ZW = CD - 2WC = 650 - 408 = \boxed{242}$

Answer:  242