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

m (Solution 2 (One Pair of Similar Triangles, then Areas): then -> Then)
m (Solution 2 (One Pair of Similar Triangles, Then Areas): Varied the word choice other than "we have".)
 
(2 intermediate revisions by one other user not shown)
Line 28: Line 28:
 
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 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>
+
Since <math>\triangle ADO\sim\triangle CPO</math> by AA, we get <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
+
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 produces
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
2[ADO]&=2[BCO] \\
 
2[ADO]&=2[BCO] \\
Line 39: Line 39:
 
\end{align*}</cmath>
 
\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>
+
In <math>\triangle CPB,</math> we obtain <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
 
~MRENTHUSIASM

Latest revision as of 15:43, 22 March 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

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 (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 get \[AD=CP\cdot\frac{DO}{PO}=h\cdot\frac{x}{11}.\]

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 produces \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 obtain \[h=\sqrt{43^2-33^2}=\sqrt{76\cdot10}=2\sqrt{190}\] 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

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

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

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

Invalid username
Login to AoPS