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

Revision as of 03:56, 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

Solution 2

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

Video Solution (Using Similar Triangles, Pythagorean Theorem)

https://youtu.be/gjeSGJy_ld4

~ pi_is_3.14

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 18: / Line 18: @@
 ~mn28407
+==Solution 2==
+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 of <math>\triangle BCD.</math> Let <math>DO=x</math> and <math>CP=h.</math> We have <math>PB=x+11.</math>
+Since <math>\triangle ADO\sim\triangle CPO</math> by AA, we have <cmath>AD=CP\cdot\frac{DO}{PO}=h\cdot\frac{x}{11}.</cmath>
+Let the brackets denote areas. Notice that <math>[ADO]=[BCO]</math> (By the same base/height, <math>[ADC]=[BCD].</math> Subtracting <math>[OCD]</math> from both sides gives <math>[ADO]=[BCO].</math>). Doubling both sides, we have
+<cmath>\begin{align*}
+[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*}</cmath>
+In <math>\triangle CPB,</math> we have <cmath>h=\sqrt{43^2-33^2}=\sqrt{76\cdot10}=2\sqrt{190}</cmath> and <cmath>AD=h\cdot\frac{x}{11}=4\sqrt{190}.</cmath> Finally, <math>4+190=\boxed{\textbf{(D) }194}.</math>
+~MRENTHUSIASM
 == Video Solution (Using Similar Triangles, Pythagorean Theorem) ==

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