2021 AMC 12A Problems/Problem 17

The following problem is from both the 2021 AMC 10A #17 and 2021 AMC 12A #17, so both problems redirect to this page.

Problem

Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD},BC=CD=43$, and $\overline{AD}\perp\overline{BD}$. Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$, and let $P$ be the midpoint of $\overline{BD}$. Given that $OP=11$, the length of $AD$ can be written in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m+n$?

$\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215$

Diagram

[asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, O, P; C = (43,0); D = (0,0); B = intersectionpoints(Circle(C,43),Circle(D,66))[0]; A = intersectionpoints(Circle(D,4*sqrt(190)),B--B+100*dir(180))[1]; P = midpoint(B--D); O = intersectionpoint(A--C,B--D); dot("$C$",C,1.5*SE,linewidth(4)); dot("$D$",D,1.5*SW,linewidth(4)); dot("$B$",B,1.5*NE,linewidth(4)); dot("$A$",A,1.5*NW,linewidth(4)); dot("$P$",P,1.5*N,linewidth(4)); dot("$O$",O,1.5*S,linewidth(4)); markscalefactor=0.25; draw(rightanglemark(A,D,O),red); draw(A--B--C--D--cycle^^A--C^^B--D^^C--P); label("$43$",B--C,E); label("$43$",C--D,S); label("$11$",midpoint(O--P),NW); [/asy] ~MRENTHUSIASM

Solution 1 (Similar Triangles and Pythagorean Theorem)

Angle chasing* reveals that $\triangle BPC\sim\triangle BDA$, therefore \[2=\frac{BD}{BP}=\frac{AB}{BC}=\frac{AB}{43},\] or $AB=86$.

Additional angle chasing shows that $\triangle ABO\sim\triangle CDO$, therefore \[2=\frac{AB}{CD}=\frac{BO}{OD}=\frac{BP+11}{BP-11},\] or $BP=33$ and $BD=66$.

Since $\triangle ADB$ is right, the Pythagorean theorem implies that \[AD=\sqrt{86^2-66^2}=4\sqrt{190}.\] The answer is $4+190=\boxed{\textbf{(D) }194}$.

  • Angle Chasing: If we set $\angle DBC = \alpha$, then we know that $\angle DCB = 180^\circ-2\alpha$ because $\triangle DBC$ is isosceles. And because $\overline{AB}\parallel\overline{DC}$, we conclude that $\angle ABD = \alpha$ too. Lastly, because $\triangle BPC$ and $\triangle BDA$ are both right triangles, they are similar by AA.

~mn28407 (Solution)

~mm (Angle Chasing Remark)

~eagleye ~MRENTHUSIASM (Minor Edits)

Solution 2 (Similar Triangles, Areas, Pythagorean Theorem)

Since $\triangle BCD$ is isosceles with base $\overline{BD},$ it follows that median $\overline{CP}$ is also an altitude. Let $OD=x$ and $CP=h,$ so $PB=x+11.$

Since $\angle AOD=\angle COP$ by vertical angles, we conclude that $\triangle AOD\sim\triangle COP$ by AA, from which $\frac{AD}{CP}=\frac{OD}{OP},$ or \[AD=CP\cdot\frac{OD}{OP}=h\cdot\frac{x}{11}.\] Let the brackets denote areas. Notice that $[AOD]=[BOC]$ (By the same base and height, we deduce that $[ACD]=[BDC].$ Subtracting $[OCD]$ from both sides gives $[AOD]=[BOC].$). Doubling both sides produces \begin{align*} 2[AOD]&=2[BOC] \\ OD\cdot AD&=OB\cdot CP \\ x\left(\frac{hx}{11}\right)&=(x+22)h \\ x^2&=11(x+22). \end{align*} Rearranging and factoring result in $(x-22)(x+11)=0,$ from which $x=22.$

Applying the Pythagorean Theorem to right $\triangle CPB,$ we have \[h=\sqrt{43^2-33^2}=\sqrt{(43+33)(43-33)}=\sqrt{760}=2\sqrt{190}.\] Finally, we get \[AD=h\cdot\frac{x}{11}=4\sqrt{190},\] so the answer is $4+190=\boxed{\textbf{(D) }194}.$

~MRENTHUSIASM

Solution 3 (Short)

Let $CP = y$. $CP$ a is perpendicular bisector of $DB.$ Then, let $DO = x,$ thus $DP = PB = 11+x.$

(1) $\triangle CPO \sim \triangle ADO,$ so we get $\frac{AD}{x} = \frac{y}{11},$ or $AD = \frac{xy}{11}.$

(2) Applying Pythagorean Theorem on $\triangle CDP$ gives $(11+x)^2 + y^2 = 43^2.$

(3) $\triangle BPC \sim \triangle BDA$ with ratio $1:2,$ so $AD = 2y$ using the fact that $P$ is the midpoint of $BD$.

Thus, $\frac{xy}{11} = 2y,$ or $x = 22.$ And $y = \sqrt{43^2 - 33^2} = 2 \sqrt{190},$ so $AD = 4 \sqrt{190}$ and the answer is $4+190=\boxed{\textbf{(D) }194}.$

~ ccx09

Solution 4 (Extending the Line)

Observe that $\triangle BPC$ is congruent to $\triangle DPC$; both are similar to $\triangle BDA$. Let's extend $\overline{AD}$ and $\overline{BC}$ past points $D$ and $C$ respectively, such that they intersect at a point $E$. Observe that $\angle BDE$ is $90$ degrees, and that $\angle DBE \cong \angle PBC \cong \angle DBA \implies \angle DBE \cong \angle DBA$. Thus, by ASA, we know that $\triangle ABD \cong \triangle EBD$, thus, $AD = ED$, meaning $D$ is the midpoint of $AE$. Let $M$ be the midpoint of $\overline{DE}$. Note that $\triangle CME$ is congruent to $\triangle BPC$, thus $BC = CE$, meaning $C$ is the midpoint of $\overline{BE}.$

Therefore, $\overline{AC}$ and $\overline{BD}$ are both medians of $\triangle ABE$. This means that $O$ is the centroid of $\triangle ABE$; therefore, because the centroid divides the median in a 2:1 ratio, $\frac{BO}{2} = DO = \frac{BD}{3}$. Recall that $P$ is the midpoint of $BD$; $DP = \frac{BD}{2}$. The question tells us that $OP = 11$; $DP-DO=11$; we can write this in terms of $DB$; $\frac{DB}{2}-\frac{DB}{3} = \frac{DB}{6} = 11 \implies DB = 66$.

We are almost finished. Each side length of $\triangle ABD$ is twice as long as the corresponding side length $\triangle CBP$ or $\triangle CPD$, since those triangles are similar; this means that $AB = 2 \cdot 43 = 86$. Now, by Pythagorean theorem on $\triangle ABD$, $AB^{2} - BD^{2} = AD^{2} \implies 86^{2}-66^{2} = AD^{2} \implies AD = \sqrt{3040} \implies AD = 4 \sqrt{190}$.

The answer is $4+190 = \boxed{\textbf{(D) }194}$.

~ ihatemath123

Solution 5

Since $P$ is the midpoint of isosceles triangle $BCD$, it would be pretty easy to see that $CP\perp BD$. Since $AD\perp BD$ as well, $AD\parallel CP$. Connecting $AP$, it’s obvious that $[ADC]=[ADP]$. Since $DP=BP$, $[APB]=[ADC]$.

Since $P$ is the midpoint of $BD$, the height of $\triangle APB$ on side $AB$ is half that of $\triangle ADC$ on $CD$. Since $[APB]=[ADC]$, $AB=2CD$.

As a basic property of a trapezoid, $\triangle AOB \sim \triangle COD$, so $\frac{OB}{OD}=\frac{AB}{CD}=2$, or $OB=2OD$. Letting $OD=x$, then $PB=DP=11+x$, and $OB=22+x$. Hence $22+x=2x$ and $x=22$.

Since $\triangle AOD \sim \triangle COP$, $\frac{AD}{PC}=\frac{OD}{OP}=2$. Since $PD=11+22=33$, $PC=\sqrt{43^2-33^2}=\sqrt{760}$.

So, $AD=2\sqrt{760}=4\sqrt{190}$. The correct answer is $\boxed{\textbf{(D) }194}$

Solution 6 (Coordinate Geometry)

Let $D$ be the origin of the cartesian coordinate plane, $B$ lie on the positive $x$-axis, and $A$ lie on the negative $y$-axis. Then let the coordinates of $B = (2a,0), A = (0, -2b).$ Then the slope of $AB$ is $\frac{b}{a}.$ Since $AB \parallel CD$ the slope of $CD$ is the same. Note that as $\triangle DCB$ is isosceles $C$ lies on $x = a.$ Thus since $CD$ has equation $y = \frac{b}{a}x$ ($D$ is the origin), $C = (a,b).$ Therefore $AC$ has equation $y = \frac{3b}{a}x - 2b$ and intersects $BD$ ($x$-axis) at $O =\left(\frac{2}{3}a, 0\right).$ The midpoint of $BD$ is $P = (a,0),$ so $OP = \frac{a}{3} = 11,$ from which $a = 33.$ Then by Pythagorean theorem on $\triangle DPC$ ($\triangle DBC$ is isosceles), we have $b = \sqrt{43^2 - 33^2} = 2\sqrt{190},$ so $2b=4\sqrt{190}.$

Finally, the answer is $4+190=\boxed{\textbf{(D) }194}.$

~Aaryabhatta1

Video Solution (Using Similar Triangles, Pythagorean Theorem)

https://youtu.be/gjeSGJy_ld4

~ pi_is_3.14

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=rtdovluzgQs

Video Solution by Mathematical Dexterity

https://www.youtube.com/watch?v=QzAVdsgBBqg

See also

2021 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2021 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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