Difference between revisions of "2021 AMC 12A Problems/Problem 17"

Revision as of 04:15, 24 April 2021

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

File:2021 AMC 12A Problem 17 (Revised).png

~MRENTHUSIASM (by Geometry Expressions)

Solution 1

Angle chasing reveals that $\triangle BPC\sim\triangle BDA$ , therefore $2=\frac{BD}{BP}=\frac{AB}{BC}=\frac{AB}{43}$ $AB=86$ Additional angle chasing shows that $\triangle ABO \sim\triangle CDO$ , therefore $2=\frac{AB}{CD}=\frac{BP}{PD}=\frac{\frac{BD}{2}+11}{\frac{BD}{2}-11}$ $BD=66$ Since $\triangle ADB$ is right, the Pythagorean theorem implies that $AD=\sqrt{86^2-66^2}$ $AD=4\sqrt{190}$ $4\sqrt{190}\implies 4 + 190 = \boxed{\textbf{D) } 194}$

~mn28407

Solution 2 (Similar Triangles, Areas, Pythagorean Theorem)

Since $\triangle BCD$ is isosceles with legs $\overline{CB}$ and $\overline{CD},$ it follows that the median $\overline{CP}$ is also an altitude. Let $OD=x$ and $CP=h.$ We have $PB=x+11.$

Since $\angle AOD=\angle COP$ by vertical angles, we get $\triangle AOD\sim\triangle COP$ 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 $[ADO]=[BCO]$ (By the same base/height, $[ADC]=[BCD].$ Subtracting $[OCD]$ from both sides gives $[ADO]=[BCO].$ ). Doubling both sides produces $\begin{align*} 2[ADO]&=2[BCO] \\ DO\cdot AD&=OB\cdot CP \\ x\left(\frac{hx}{11}\right)&=(x+22)h \\ x^2&=11(x+22). \end{align*}$ Rearranging and factoring result $(x-22)(x+11)=0,$ from which $x=22.$

Applying the Pythagorean Theorem to right $\triangle CPB,$ we obtain $h=\sqrt{43^2-33^2}=\sqrt{(43+33)(43-33)}=\sqrt{760}=2\sqrt{190}.$

Finally, we conclude that $AD=h\cdot\frac{x}{11}=4\sqrt{190},$ so the answer is $4+190=\boxed{\textbf{(D) }194}.$

~MRENTHUSIASM

Solution 3 (Short)

Let $CP = y$ and $CP$ is perpendicular bisector of $DB.$ Let $DO = x,$ so $DP = PB = 11+x.$

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

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

(3) $\triangle BPC \sim \triangle BDA$ with ratio $1:2,$ so $AD = 2y.$

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

~ 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 $\triangle ABD$ , $AB^{2} - BD^{2} = AD^{2} \implies 86^{2}-66^{2} = AD^{2} \implies AD = \sqrt{3040} \implies AD = 4 \sqrt{190}$ . $4+190 = \boxed{194, \textbf{D}}$

~ ihatemath123

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

@@ Line 26: / Line 26: @@
 ==Solution 2 (Similar Triangles, Areas, Pythagorean Theorem)==
-Since <math>\triangle BCD</math> is isosceles with legs <math>\overline{CB}</math> and <math>\overline{CD},</math> it follows that the median <math>\overline{CP}</math> is also an altitude. Let <math>DO=x</math> and <math>CP=h.</math> We have <math>PB=x+11.</math>
+Since <math>\triangle BCD</math> is isosceles with legs <math>\overline{CB}</math> and <math>\overline{CD},</math> it follows that the median <math>\overline{CP}</math> is also an altitude. Let <math>OD=x</math> and <math>CP=h.</math> We have <math>PB=x+11.</math>
 Since <math>\angle AOD=\angle COP</math> by vertical angles, we get <math>\triangle AOD\sim\triangle COP</math> by AA, from which <math>\frac{AD}{CP}=\frac{OD}{OP},</math> or <cmath>AD=CP\cdot\frac{OD}{OP}=h\cdot\frac{x}{11}.</cmath>

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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All AMC 10 Problems and Solutions

2021 AMC 12A (Problems • Answer Key • Resources)
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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