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Difference between revisions of "2022 AIME I Problems/Problem 3"
Line 36: Line 36:
 
</asy>
 
</asy>
  
−
== Solution 1==
+
== Solution 1 ==
 +
 
 +
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>
 +
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 P = (-121, 162.2);
 +
pair P' =(-412.5,162.2);
 +
pair Q = (121, 162.2);
 +
pair Q' = (412.5,162.5);
 +
dot(P);
 +
dot(Q);
 +
 
 +
draw(P'--Q');
 +
 
 +
label("$A$", A, N);
 +
label("$B$", B, N);
 +
 
 +
label("$C$", C, S);
 +
label("$D$", D, S);
 +
 
 +
label("$P$", P, N);
 +
label("$Q$", Q, N);
 +
 
 +
label("$P'$", P', W);
 +
label($"Q'$", Q', E);
 +
</asy>
 +
 
 +
 
 +
== Solution 2==
  
 
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.

Revision as of 19:10, 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: 

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 P = (-121, 162.2);
pair P' =(-412.5,162.2);
pair Q = (121, 162.2);
pair Q' = (412.5,162.5);
dot(P);
dot(Q);

draw(P'--Q');

label("$A$", A, N);
label("$B$", B, N);

label("$C$", C, S);
label("$D$", D, S);

label("$P$", P, N);
label("$Q$", Q, N);

label("$P'$", P', W);
label($"Q'$", Q', E);
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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

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

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