AIME 2022 I · 第 3 题

Solution 1

We have the following diagram:

Let $X$ and $W$ be the points where $AP$ and $BQ$ extend to meet $CD$, and $YZ$ be the height of $\triangle AZB$. As proven in Solution 2, triangles $APD$ and $DPW$ are congruent right triangles. Therefore, $AD = DW = 333$. We can apply this logic to triangles $BCQ$ and $XCQ$ as well, giving us $BC = CX = 333$. Since $CD = 650$, $XW = DW + CX - CD = 16$.

Additionally, we can see that $\triangle XZW$ is similar to $\triangle PQZ$ and $\triangle AZB$. We know that $\frac{XW}{AB} = \frac{16}{500}$. So, we can say that the height of the triangle $AZB$ is $500u$ while the height of the triangle $XZW$ is $16u$. After that, we can figure out the distance from $Y$ to $PQ: \frac{500+16}{2} = 258u$ and the height of triangle $PZQ: 500-258 = 242u$.

Finally, since the ratio between the height of $PZQ$ to the height of $AZB$ is $242:500$ and $AB$ is $500$, $PQ = \boxed{242}.$

~Cytronical

Solution 2

Extend line $PQ$ to meet $AD$ at $P'$ and $BC$ at $Q'$. The diagram looks like this:

Because the trapezoid is isosceles, by symmetry $PQ$ is parallel to $AB$ and $CD$. 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 3

We have the following diagram:

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 $BYXC$ 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 4

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 $PQ = ZW = CD - 2WC = 650 - 408 = \boxed{242}$.

~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 $P$ is equidistant from $AB, AD,$ and $CD$ and $Q$ is equidistant from $AB, BC,$ and $CD.$ If we let the feet of the altitudes from $P$ to $AB, AD,$ and $CD$ be called $E, F,$ and $G$ respectively, we can say that $PE = PF = PG.$ Analogously, we let the feet of the altitudes from $Q$ to $AB, BC,$ and $CD$ be $H, I,$ and $J$ respectively. Thus, $QH = QI = QJ.$ Because $ABCD$ 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 $AEP, AFP, BHQ,$ and $BIQ$ are congruent. Similarly, $DFP, DGP, CJQ,$ and $CIQ$ by the same reasoning. Additionally, $EH = GJ = PQ$ since $EHQP$ and $GJQP$ are congruent rectangles.

If we then let $x = AE = AF = BH = BI,$ let $y = CI = CJ = DG = DF,$ and let $z = EH = GJ = PQ,$ we can create the following system of equations with the given side length information:

$$\begin{aligned} 2x + z &= 500, \\ 2y + z &= 650, \\ x + y &= 333. \end{aligned}$$ Adding the first two equations, subtracting by twice the third, and dividing by $2$ yields $z = PQ = \boxed{242}.$

~regular

Solution 6

Extend line $PQ$ to meet $AD$ at $P'$ and $BC$ at $Q'$. The diagram looks like this:

Since $\angle A + \angle D=\angle B + \angle C = 180^{\circ}$, it follows that $\angle P'AP+\angle P'DP = \angle Q'BQ + \angle Q'CQ = 90^{\circ}$. Thus, $\angle APD = \angle BQC = 90^{\circ}$, implying that $\triangle APD$ and $\triangle BQC$ are right triangles. Since $P'P$ and $Q'Q$ are medians, $P'P+Q'Q=\frac{333\times2}{2}=333$. Since $P'Q'=\frac{500+650}{2}=575$, we have $PQ+P'P+Q'Q=575$, or $PQ=575-333=\boxed{242}$.

~sigma

Solution 7 (Trigonometry)

Let $PQ = x$. Note that since $AP$ bisects $\angle{A}$ and $DP$ bisects $\angle{D}$, we have

$$\angle{APD} = 180^{\circ}-\tfrac12 \angle{A}-\tfrac12 \angle{D}=90^{\circ}.$$ Let $\angle{ADP}=\theta$. We have that $\angle{ADC} = 2\theta.$ Now, drop an altitude from $A$ to $CD$ at $E$. Notice that $DE=\tfrac{650-500}{2}=75$. By the definition of cosine, we have

$$\cos{2\theta}=1-2\cos^2{\theta}=\tfrac{75}{333}=\tfrac{25}{111} \implies \cos{\theta}=\tfrac{2\sqrt{1887}}{111}.$$ Notice, however, that we can also apply this to $\triangle{APD}$; we have

$$\cos{\theta}=\tfrac{DP}{333} \implies DP=6\sqrt{1887}.$$ By the Pythagorean Theorem, we get

$$AP=\sqrt{333^2-(6\sqrt{1887})^2}=3\sqrt{4773}.$$ Then, drop an altitude from $P$ to $AB$ at $F$; if $AF=y$, then $PQ=x=500-2y$. Because $AP$ is an angle bisector, we see that $\angle{BAP}=\angle{DAP}=90^{\circ}-\theta$. Again, by the definition of cosine, we have

$$\cos{(90^{\circ}-\theta)}=\sin{\theta}=\tfrac{\sqrt{4773}}{111}=\tfrac{y}{3\sqrt{4773}} \implies y=129.$$ Finally, $PQ=500-2y=\boxed{242}$.

~pqr.

Solution 8 (Pythagoras + Similar Triangles)

As in solution 4, $\angle APD = 90^{\circ}$. Set $k = AX$ and $x = DP$.

We know that $DZ = AX + \frac{DC-AB}{2}$, so $DZ = k + \frac{650-500}{2} = k + 75$.

$\triangle DPZ \sim \triangle APD$ by AA, so we have $\frac{PD}{AD} = \frac{ZD}{PD}$, resulting in

$$\frac{x}{333} = \frac{k+75}{x} \text{ (1)}$$ $\triangle APX \sim \triangle ADP$ by AA, so we have $\frac{AP}{AD} = \frac{AX}{AP}$, resulting in

$$\frac{\sqrt{333^2-x^2}}{333} = \frac{k}{\sqrt{333^2-k^2}} \text{ (2)}$$ From $\text{(1)}$, we have $x^2 = 333k + 333(75) = 333k + 24975$. From $\text{(2)}$, we have $333^2 - x^2 = 333k$, or $x^2 = 333^2 - 333k$. Thus, $333k + 24975 = 333^2 - 333k$. Solving for $k$ yields $k = 129$.

By symmetry, $YB = AX = 129$. Thus, $PQ = XY = AB - 2AX = 500 - 2(129) = \boxed{242}$.

~ adam_zheng

Solution 9 (Really Quick)

Extend $PQ$. Let the endpoints of this extension be $X$ and $Y$, ($X$ is closer to $P$ and $Y$ is closer to $Q$). Now let $PX = QY = a$. So, $PQ = 575-2a$. Now, by transversals, $\angle{PAB} = \angle{XPA}$. So, $\triangle{XAP}$ is isosceles with $XA = XP = a$. Now, lets look at $XD$. $XD = 333-a$. Also, by transversals, $\angle{DPX} = \angle{PDC}$. Thus, we have $XD = XP$. Aha! So, $333-a = a \Longrightarrow a = \dfrac{333}{2}$. Plugging this into our equation for $PQ$, we get $575-333 = \boxed{242}$

~jb2015007

Solution 10 (Fast Trigonometric Solve)

Noticing immediately that $\Delta APD$ and $\Delta BQC$ are right triangles due to the angle bisector conditions, we are motivated to use trigonometry and half-angles. An obvious first step is to compute the cosine of $\angle{D}$, which we can drop the altitude from $A$ and find $\cos(\angle ADC) = \frac{25}{111}$. We can easily then see that by half angles, $\cos(\angle ADP) = \cos(\angle PDC) = \sqrt{\frac{68}{111}}$.

Motivated by right triangle trigonometry, we drop the altitude from $P$ to $CD$ to form point $X$. By our derived cosines, $\frac{DP}{AD} = \frac{DE}{PD} = \sqrt{\frac{68}{111}}$, so we can do some mathematics to find $DE = AD \cdot \frac{68}{111} = \frac{333 \cdot 68}{111} = 204$. To find $PQ$, we can drop the altitude from $Q$ similarly to the construction of $X$ to form point $Y$, and $PQ = XY = CD - DX - CY = CD = 2DX = 650-2\cdot204 = \boxed{242}$

~Tiguhbabheow

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

https://youtu.be/Q_S_VhiLRJE

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