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

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(→‎Solution 2 (One Pair of Similar Triangles, then Areas))
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~MRENTHUSIASM
 
~MRENTHUSIASM
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==Solution 3 (short)==
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Let <math>CP = y</math> and <math>CP</math> is perpendicular bisector of <math>DB.</math> Let <math>DO = x,</math> so <math>DP = PB = 11+x.</math>
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(1) <math>\triangle CPO \sim \triangle ADO,</math> so we get <math>\frac{AD}{x} = \frac{y}{11},</math> or <math>AD = \frac{xy}{11}.</math>
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(2) pythag on <math>\triangle CDP</math> gives <math>(11+x)^2 + y^2 = 43^2.</math>
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(3) <math>\triangle BPC \sim \triangle BDA</math> with ratio <math>1:2,</math> so <math>AD = 2y.</math>
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Thus, <math>xy/11 = 2y,</math> or <math>x = 22.</math> And <math>y = \sqrt{43^2 - 33^2} = 2 \sqrt{190},</math> so <math>AD = 4 \sqrt{190}</math> and the answer is <math>\boxed{194}.</math>
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~ ccx09
  
 
== Video Solution (Using Similar Triangles, Pythagorean Theorem) ==
 
== Video Solution (Using Similar Triangles, Pythagorean Theorem) ==

Revision as of 15:18, 13 February 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 $m+n$?

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

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 (One Pair of Similar Triangles, then Areas)

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

Since $\triangle ADO\sim\triangle CPO$ by AA, we have \[AD=CP\cdot\frac{DO}{PO}=h\cdot\frac{x}{11}.\]

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, we have \begin{align*} 2[ADO]&=2[BCO] \\ \frac{x^2 h}{11}&=(x+22)h \\ x^2&=11x+11\cdot22 \\ (x-22)(x+11)&=0 \\ x&=22. \end{align*}

In $\triangle CPB,$ we have \[h=\sqrt{43^2-33^2}=\sqrt{76\cdot10}=2\sqrt{190}\] and \[AD=h\cdot\frac{x}{11}=4\sqrt{190}.\] Finally, $4+190=\boxed{\textbf{(D) }194}.$

~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

Video Solution (Using Similar Triangles, Pythagorean Theorem)

https://youtu.be/gjeSGJy_ld4

~ pi_is_3.14

See also

2021 AMC 10A (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 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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