Difference between revisions of "2022 AIME I Problems/Problem 3"
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== Diagram == | == Diagram == | ||
<asy> | <asy> | ||
− | + | /* Made by MRENTHUSIASM */ | |
− | pair A = (-250, | + | size(300); |
− | + | pair A, B, C, D, A1, B1, C1, D1, P, Q; | |
− | + | A = (-250,6*sqrt(731)); | |
− | + | B = (250,6*sqrt(731)); | |
− | draw(A--B--C--D--cycle); | + | C = (325,-6*sqrt(731)); |
− | pair W = ( | + | 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 == | ||
+ | We have the following diagram: | ||
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM , modified by Cytronical */ | ||
+ | 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 = (0,6*sqrt(731)); | ||
+ | Z = intersectionpoint(A--4*(P-A)+A,B--4*(Q-B)+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,4.5*dir(75),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--W^^Q--X^^Y--Z,dashed); | ||
+ | </asy> | ||
− | + | Let <math>X</math> and <math>W</math> be the points where <math>AP</math> and <math>BQ</math> extend to meet <math>CD</math>, and <math>YZ</math> be the height of <math>\triangle AZB</math>. As proven in Solution 2, triangles <math>APD</math> and <math>DPW</math> are congruent right triangles. Therefore, <math>AD = DW = 333</math>. We can apply this logic to triangles <math>BCQ</math> and <math>XCQ</math> as well, giving us <math>BC = CX = 333</math>. Since <math>CD = 650</math>, <math>XW = DW + CX - CD = 16</math>. | |
− | |||
− | |||
− | |||
− | + | Additionally, we can see that <math>\triangle XZW</math> is similar to <math>\triangle PQZ</math> and <math>\triangle AZB</math>. We know that <math>\frac{XW}{AB} = \frac{16}{500}</math>. So, we can say that the height of the triangle <math>AZB</math> is <math>500u</math> while the height of the triangle <math>XZW</math> is <math>16u</math>. After that, we can figure out the distance from <math>Y</math> to <math>PQ: \frac{500+16}{2} = 258u</math> and the height of triangle <math>PZQ: 500-258 = 242u</math>. | |
− | + | ||
− | + | Finally, since the ratio between the height of <math>PZQ</math> to the height of <math>AZB</math> is <math>242:500</math> and <math>AB</math> is <math>500</math>, <math>PQ = \boxed{242}.</math> | |
− | draw(D | + | |
− | + | ~Cytronical | |
− | + | ||
− | + | == Solution 2 == | |
− | + | ||
− | + | Extend line <math>PQ</math> to meet <math>AD</math> at <math>P'</math> and <math>BC</math> at <math>Q'</math>. 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> | </asy> | ||
+ | Because the trapezoid is isosceles, by symmetry <math>PQ</math> is parallel to <math>AB</math> and <math>CD</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>. | ||
− | == | + | 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>. |
+ | ~KingRavi | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | 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 <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. | 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. | ||
− | Claim: quadrilaterals <math>AZWD</math> and <math> | + | Claim: quadrilaterals <math>AZWD</math> and <math>BYXC</math> are rhombuses. |
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. | 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. | ||
Line 47: | Line 216: | ||
~ihatemath123 | ~ihatemath123 | ||
+ | |||
+ | == Solution 4 == | ||
+ | |||
+ | <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 <math>X</math> and <math>Y</math> be the feet of the altitudes from <math>P</math> and <math>Q</math>, respectively, to <math>AB</math>, and let <math>Z</math> and <math>W</math> be the feet of the altitudes from <math>P</math> and <math>Q</math>, respectively, to <math>CD</math>. Side <math>AB</math> is parallel to side <math>CD</math>, so <math>XYWZ</math> is a rectangle with width <math>PQ</math>. Furthermore, because <math>CD - AB = 650-500 = 150</math> and trapezoid <math>ABCD</math> is isosceles, <math>WC - YB = ZD - XA = 75</math>. | ||
+ | |||
+ | Also because <math>ABCD</math> is isosceles, <math>\angle ABC + \angle BCD</math> is half the total sum of angles in <math>ABCD</math>, or <math>180^{\circ}</math>. Since <math>BQ</math> and <math>CQ</math> bisect <math>\angle ABC</math> and <math>\angle BCD</math>, respectively, we have <math>\angle QBC + \angle QCB = 90^{\circ}</math>, so <math>\angle BQC = 90^{\circ}</math>. | ||
+ | |||
+ | Letting <math>BQ = 333k</math>, applying Pythagoras to <math>\triangle BQC</math> yields <math>QC = 333\sqrt{1-k^2}</math>. We then proceed using similar triangles: <math>\angle BYQ = \angle BQC = 90^{\circ}</math> and <math>\angle YBQ = \angle QBC</math>, so by AA similarity <math>YB = 333k^2</math>. Likewise, <math>\angle CWQ = \angle BQC = 90^{\circ}</math> and <math>\angle WCQ = \angle QCB</math>, so by AA similarity <math>WC = 333(1 - k^2)</math>. Thus <math>WC + YB = 333</math>. | ||
+ | |||
+ | Adding our two equations for <math>WC</math> and <math>YB</math> gives <math>2WC = 75 + 333 = 408</math>. Therefore, the answer is <math>PQ = ZW = CD - 2WC = 650 - 408 = \boxed{242}</math>. | ||
+ | |||
+ | ~Orange_Quail_9 | ||
+ | |||
+ | == Solution 5 == | ||
+ | |||
+ | This will be my first solution on AoPS. My apologies in advance for any errors. | ||
+ | |||
+ | Angle bisectors can be thought of as the locus of all points equidistant from the lines whose angle they bisect. It can thus be seen that <math>P</math> is equidistant from <math>AB, AD,</math> and <math>CD</math> and <math>Q</math> is equidistant from <math>AB, BC,</math> and <math>CD.</math> If we let the feet of the altitudes from <math>P</math> to <math>AB, AD,</math> and <math>CD</math> be called <math>E, F,</math> and <math>G</math> respectively, we can say that <math>PE = PF = PG.</math> Analogously, we let the feet of the altitudes from <math>Q</math> to <math>AB, BC,</math> and <math>CD</math> be <math>H, I,</math> and <math>J</math> respectively. Thus, <math>QH = QI = QJ.</math> Because <math>ABCD</math> is an isosceles trapezoid, we can say that all of the altitudes are equal to each other. | ||
+ | |||
+ | By SA as well as SS congruence for right triangles, we find that triangles <math>AEP, AFP, BHQ,</math> and <math>BIQ</math> are congruent. Similarly, <math>DFP, DGP, CJQ,</math> and <math>CIQ</math> by the same reasoning. Additionally, <math>EH = GJ = PQ</math> since <math>EHQP</math> and <math>GJQP</math> are congruent rectangles. | ||
+ | |||
+ | If we then let <math>x = AE = AF = BH = BI,</math> let <math>y = CI = CJ = DG = DF,</math> and let <math>z = EH = GJ = PQ,</math> we can create the following system of equations with the given side length information: | ||
+ | <cmath>\begin{align*} | ||
+ | 2x + z &= 500, \\ | ||
+ | 2y + z &= 650, \\ | ||
+ | x + y &= 333. | ||
+ | \end{align*}</cmath> | ||
+ | Adding the first two equations, subtracting by twice the second, and dividing by <math>2</math> yields <math>z = PQ = \boxed{242}.</math> | ||
+ | |||
+ | ~regular | ||
+ | |||
+ | == Solution 6 == | ||
+ | Extend line <math>PQ</math> to meet <math>AD</math> at <math>P'</math> and <math>BC</math> at <math>Q'</math>. 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> | ||
+ | Since <math>\angle A + \angle D=\angle B + \angle C = 180^{\circ}</math>, it follows that <math>\angle P'AP+\angle P'DP = \angle Q'BQ + \angle Q'CQ = 90^{\circ}</math>. Thus, <math>\angle APD = \angle BQC = 90^{\circ}</math>, implying that <math>\triangle APD</math> and <math>\triangle BQC</math> are right triangles. Since <math>P'P</math> and <math>Q'Q</math> are medians, <math>P'P+Q'Q=\frac{333\times2}{2}=333</math>. Since <math>P'Q'=\frac{500+650}{2}=575</math>, we have <math>PQ+P'P+Q'Q=575</math>, or <math>PQ=575-333=\boxed{242}</math>. | ||
+ | |||
+ | ~sigma | ||
+ | |||
+ | == Solution 7 (Trigonometry) == | ||
+ | |||
+ | Let <math>PQ = x</math>. Note that since <math>AP</math> bisects <math>\angle{A}</math> and <math>DP</math> bisects <math>\angle{D}</math>, we have <cmath>\angle{APD} = 180^{\circ}-\tfrac12 \angle{A}-\tfrac12 \angle{D}=90^{\circ}.</cmath> Let <math>\angle{ADP}=\theta</math>. We have that <math>\angle{ADC} = 2\theta.</math> Now, drop an altitude from <math>A</math> to <math>CD</math> at <math>E</math>. Notice that <math>DE=\tfrac{650-500}{2}=75</math>. By the definition of cosine, we have <cmath>\cos{2\theta}=1-2\cos^2{\theta}=\tfrac{75}{333}=\tfrac{25}{111} \implies \cos{\theta}=\tfrac{2\sqrt{1887}}{111}.</cmath> Notice, however, that we can also apply this to <math>\triangle{APD}</math>; we have <cmath>\cos{\theta}=\tfrac{DP}{333} \implies DP=6\sqrt{1887}.</cmath> By the Pythagorean Theorem, we get <cmath>AP=\sqrt{333^2-(6\sqrt{1887})^2}=3\sqrt{4773}.</cmath> Then, drop an altitude from <math>P</math> to <math>AB</math> at <math>F</math>; if <math>AF=y</math>, then <math>PQ=x=500-2y</math>. Because <math>AP</math> is an angle bisector, we see that <math>\angle{BAP}=\angle{DAP}=90^{\circ}-\theta</math>. Again, by the definition of cosine, we have <cmath>\cos{(90^{\circ}-\theta)}=\sin{\theta}=\tfrac{\sqrt{4773}}{111}=\tfrac{y}{3\sqrt{4773}} \implies y=129.</cmath> Finally, <math>PQ=500-2y=\boxed{242}</math>. | ||
+ | |||
+ | ~pqr. | ||
+ | |||
+ | == Solution 8 (Pythagoras + Similar Triangles) == | ||
+ | <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> | ||
+ | |||
+ | As in solution 4, <math>\angle APD = 90^{\circ}</math>. Set <math>k = AX</math> and <math>x = DP</math>. | ||
+ | |||
+ | We know that <math>DZ = AX + \frac{DC-AB}{2}</math>, so <math>DZ = k + \frac{650-500}{2} = k + 75</math>. | ||
+ | |||
+ | <math>\triangle DPZ \sim \triangle APD</math> by AA, so we have <math>\frac{PD}{AD} = \frac{ZD}{PD}</math>, resulting in | ||
+ | |||
+ | <cmath> | ||
+ | \frac{x}{333} = \frac{k+75}{x} \text{ (1)} | ||
+ | </cmath> | ||
+ | |||
+ | <math>\triangle APX \sim \triangle ADP</math> by AA, so we have <math>\frac{AP}{AD} = \frac{AX}{AP}</math>, resulting in | ||
+ | |||
+ | <cmath> | ||
+ | \frac{\sqrt{333^2-x^2}}{333} = \frac{k}{\sqrt{333^2-k^2}} \text{ (2)} | ||
+ | </cmath> | ||
+ | |||
+ | From <math>\text{(1)}</math>, we have <math>x^2 = 333k + 333(75) = 333k + 24975</math>. From <math>\text{(2)}</math>, we have <math>333^2 - x^2 = 333k</math>, or <math>x^2 = 333^2 - 333k</math>. Thus, <math>333k + 24975 = 333^2 - 333k</math>. Solving for <math>k</math> yields <math>k = 129</math>. | ||
+ | |||
+ | By symmetry, <math>YB = AX = 129</math>. Thus, <math>PQ = XY = AB - 2AX = 500 - 2(129) = \boxed{242}</math>. | ||
+ | |||
+ | ~ adam_zheng | ||
==Video Solution (Mathematical Dexterity)== | ==Video Solution (Mathematical Dexterity)== | ||
https://www.youtube.com/watch?v=fNAvxXnvAxs | https://www.youtube.com/watch?v=fNAvxXnvAxs | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=h_LOT-rwt08 | ||
+ | |||
+ | ~Steven Chen (www.professorchenedu.com) | ||
+ | |||
+ | == Video Solution == | ||
+ | |||
+ | https://youtu.be/MJ_M-xvwHLk?t=545 | ||
+ | |||
+ | ~ThePuzzlr | ||
+ | |||
+ | ==Video Solution by MRENTHUSIASM (English & Chinese)== | ||
+ | https://www.youtube.com/watch?v=dqVVOSCWujo&ab_channel=MRENTHUSIASM | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/Q_S_VhiLRJE | ||
+ | |||
+ | ~AMC & AIME Training | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2022|n=I|num-b=2|num-a=4}} | {{AIME box|year=2022|n=I|num-b=2|num-a=4}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:59, 2 March 2024
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1
- 4 Solution 2
- 5 Solution 3
- 6 Solution 4
- 7 Solution 5
- 8 Solution 6
- 9 Solution 7 (Trigonometry)
- 10 Solution 8 (Pythagoras + Similar Triangles)
- 11 Video Solution (Mathematical Dexterity)
- 12 Video Solution
- 13 Video Solution
- 14 Video Solution by MRENTHUSIASM (English & Chinese)
- 15 Video Solution
- 16 See Also
Problem
In isosceles trapezoid , parallel bases
and
have lengths
and
, respectively, and
. The angle bisectors of
and
meet at
, and the angle bisectors of
and
meet at
. Find
.
Diagram
~MRENTHUSIASM ~ihatemath123
Solution 1
We have the following diagram:
Let and
be the points where
and
extend to meet
, and
be the height of
. As proven in Solution 2, triangles
and
are congruent right triangles. Therefore,
. We can apply this logic to triangles
and
as well, giving us
. Since
,
.
Additionally, we can see that is similar to
and
. We know that
. So, we can say that the height of the triangle
is
while the height of the triangle
is
. After that, we can figure out the distance from
to
and the height of triangle
.
Finally, since the ratio between the height of to the height of
is
and
is
,
~Cytronical
Solution 2
Extend line to meet
at
and
at
. The diagram looks like this:
Because the trapezoid is isosceles, by symmetry
is parallel to
and
. Therefore,
by interior angles and
by the problem statement. Thus,
is isosceles with
. By symmetry,
is also isosceles, and thus
. Similarly, the same thing is happening on the right side of the trapezoid, and thus
is the midline of the trapezoid. Then,
.
Since and
, we have
. The length of the midline of a trapezoid is the average of their bases, so
. Finally,
.
~KingRavi
Solution 3
We have the following diagram:
Extend lines
and
to meet line
at points
and
, respectively, and extend lines
and
to meet
at points
and
, respectively.
Claim: quadrilaterals and
are rhombuses.
Proof: Since ,
. Therefore, triangles
,
,
and
are all right triangles. By SAA congruence, the first three triangles are congruent; by SAS congruence,
is congruent to the other three. Therefore,
, so
is a rhombus. By symmetry,
is also a rhombus.
Extend line to meet
and
at
and
, respectively. Because of rhombus properties,
. Also, by rhombus properties,
and
are the midpoints of segments
and
, respectively; therefore, by trapezoid properties,
. Finally,
.
~ihatemath123
Solution 4
Let and
be the feet of the altitudes from
and
, respectively, to
, and let
and
be the feet of the altitudes from
and
, respectively, to
. Side
is parallel to side
, so
is a rectangle with width
. Furthermore, because
and trapezoid
is isosceles,
.
Also because is isosceles,
is half the total sum of angles in
, or
. Since
and
bisect
and
, respectively, we have
, so
.
Letting , applying Pythagoras to
yields
. We then proceed using similar triangles:
and
, so by AA similarity
. Likewise,
and
, so by AA similarity
. Thus
.
Adding our two equations for and
gives
. Therefore, the answer is
.
~Orange_Quail_9
Solution 5
This will be my first solution on AoPS. My apologies in advance for any errors.
Angle bisectors can be thought of as the locus of all points equidistant from the lines whose angle they bisect. It can thus be seen that is equidistant from
and
and
is equidistant from
and
If we let the feet of the altitudes from
to
and
be called
and
respectively, we can say that
Analogously, we let the feet of the altitudes from
to
and
be
and
respectively. Thus,
Because
is an isosceles trapezoid, we can say that all of the altitudes are equal to each other.
By SA as well as SS congruence for right triangles, we find that triangles and
are congruent. Similarly,
and
by the same reasoning. Additionally,
since
and
are congruent rectangles.
If we then let let
and let
we can create the following system of equations with the given side length information:
Adding the first two equations, subtracting by twice the second, and dividing by
yields
~regular
Solution 6
Extend line to meet
at
and
at
. The diagram looks like this:
Since
, it follows that
. Thus,
, implying that
and
are right triangles. Since
and
are medians,
. Since
, we have
, or
.
~sigma
Solution 7 (Trigonometry)
Let . Note that since
bisects
and
bisects
, we have
Let
. We have that
Now, drop an altitude from
to
at
. Notice that
. By the definition of cosine, we have
Notice, however, that we can also apply this to
; we have
By the Pythagorean Theorem, we get
Then, drop an altitude from
to
at
; if
, then
. Because
is an angle bisector, we see that
. Again, by the definition of cosine, we have
Finally,
.
~pqr.
Solution 8 (Pythagoras + Similar Triangles)
As in solution 4, . Set
and
.
We know that , so
.
by AA, so we have
, resulting in
by AA, so we have
, resulting in
From , we have
. From
, we have
, or
. Thus,
. Solving for
yields
.
By symmetry, . Thus,
.
~ adam_zheng
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)
Video Solution
https://youtu.be/MJ_M-xvwHLk?t=545
~ThePuzzlr
Video Solution by MRENTHUSIASM (English & Chinese)
https://www.youtube.com/watch?v=dqVVOSCWujo&ab_channel=MRENTHUSIASM
~MRENTHUSIASM
Video Solution
~AMC & AIME Training
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.