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2021 AMC 12A Problems/Problem 17: Difference between revisions

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Line 12: Line 12:
<cmath>2=\frac{AB}{CD}=\frac{BP}{PD}=\frac{\frac{BD}{2}+11}{\frac{BD}{2}-11}</cmath>
<cmath>2=\frac{AB}{CD}=\frac{BP}{PD}=\frac{\frac{BD}{2}+11}{\frac{BD}{2}-11}</cmath>
<cmath>BD=66</cmath>
<cmath>BD=66</cmath>
Since <math>\triangle ADB</math> is right, the Pythagorean theorem reveals that
Since <math>\triangle ADB</math> is right, the Pythagorean theorem implies that
<cmath>AD=\sqrt{86^2-66^2}</cmath>
<cmath>AD=\sqrt{86^2-66^2}</cmath>
<cmath>AD=4\sqrt{190}</cmath>
<cmath>AD=4\sqrt{190}</cmath>

Revision as of 01:11, 12 February 2021

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

Video Solution (Using Similar Triangles, Pythagorean Theorem)

https://youtu.be/gjeSGJy_ld4

~ pi_is_3.14

See also

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
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

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