# 2020 AMC 10A Problems/Problem 12

## Problem 12

Triangle $AMC$ is isosceles with $AM = AC$. Medians $\overline{MV}$ and $\overline{CU}$ are perpendicular to each other, and $MV=CU=12$. What is the area of $\triangle AMC?$

$[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); label("M", (-4,0), W); label("C", (4,0), E); label("A", (0, 12), N); label("V", (2, 6), NE); label("U", (-2, 6), NW); label("P", (0, 3.6), S); [/asy]$

$\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192$

## Solution 1

Since quadrilateral $UVCM$ has perpendicular diagonals, its area can be found as half of the product of the length of the diagonals. Also note that $\triangle AUV$ has $\frac 14$ the area of triangle $AMC$ by similarity, so $[UVCM]=\frac 34\cdot [AMC].$ Thus, $$\frac 12 \cdot 12\cdot 12=\frac 34 \cdot [AMC]$$ $$72=\frac 34\cdot [AMC]$$ $$[AMC]=96\rightarrow \boxed{\textbf{(C)}}.$$

## Solution 2 (Trapezoid)

$[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((-2,6)--(2,6)); label("M", (-4,0), W); label("C", (4,0), E); label("A", (0, 12), N); label("V", (2, 6), NE); label("U", (-2, 6), NW); label("P", (0, 3.6), S); [/asy]$

We know that $\triangle AUV \sim \triangle AMC$, and since the ratios of its sides are $\frac{1}{2}$, the ratio of of their areas is $\left(\frac{1}{2}\right)^2=\frac{1}{4}$.

If $\triangle AUV$ is $\frac{1}{4}$ the area of $\triangle AMC$, then trapezoid $MUVC$ is $\frac{3}{4}$ the area of $\triangle AMC$.

Let's call the intersection of $\overline{UC}$ and $\overline{MV}$ $P$. Let $\overline{UP}=x$. Then $\overline{PC}=12-x$. Since $\overline{UC} \perp \overline{MV}$, $\overline{UP}$ and $\overline{CP}$ are heights of triangles $\triangle MUV$ and $\triangle MCV$, respectively. Both of these triangles have base $12$.

Area of $\triangle MUV = \frac{x\cdot12}{2}=6x$

Area of $\triangle MCV = \frac{(12-x)\cdot12}{2}=72-6x$

Adding these two gives us the area of trapezoid $MUVC$, which is $6x+(72-6x)=72$.

This is $\frac{3}{4}$ of the triangle, so the area of the triangle is $\frac{4}{3}\cdot{72}=\boxed{\textbf{(C) } 96}$ ~quacker88, diagram by programjames1

## Solution 3 (Medians)

Draw median $\overline{AB}$. $[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((0,12)--(0,0)); label("M", (-4,0), W); label("C", (4,0), E); label("A", (0, 12), N); label("V", (2, 6), NE); label("U", (-2, 6), NW); label("P", (0.5, 4), E); label("B", (0, 0), S); [/asy]$

Since we know that all medians of a triangle intersect at the centroid, we know that $\overline{AB}$ passes through point $P$. We also know that medians of a triangle divide each other into segments of ratio $2:1$. Knowing this, we can see that $\overline{PC}:\overline{UP}=2:1$, and since the two segments sum to $12$, $\overline{PC}$ and $\overline{UP}$ are $8$ and $4$, respectively.

Finally knowing that the medians divide the triangle into $6$ sections of equal area, finding the area of $\triangle PUM$ is enough. $\overline{PC} = \overline{MP} = 8$.

The area of $\triangle PUM = \frac{4\cdot8}{2}=16$. Multiplying this by $6$ gives us $6\cdot16=\boxed{\textbf{(C) }96}$

~quacker88

## Solution 4 (Triangles)

$[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((-2,6)--(2,6)); label("M", (-4,0), W); label("C", (4,0), E); label("A", (0, 12), N); label("V", (2, 6), NE); label("U", (-2, 6), NW); label("P", (0, 3.6), S); [/asy]$ We know that $AU = UM$, $AV = VC$, so $UV = \frac{1}{2} MC$.

As $\angle UPM = \angle VPC = 90$, we can see that $\triangle UPM \cong \triangle VPC$ and $\triangle UVP \sim \triangle MPC$ with a side ratio of $1 : 2$.

So $UP = VP = 4$, $MP = PC = 8$.

With that, we can see that $[\triangle UPM] = 16$, and the area of trapezoid $MUVC$ is 72.

As said in solution 1, $[\triangle AMC] = 72 / \frac{3}{4} = \boxed{\textbf{(C) } 96}$.

## Solution 5 (Only Pythagorean Theorem)

$[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((0,12)--(0,0)); label("M", (-4,0), W); label("C", (4,0), E); label("A", (0, 12), N); label("V", (2, 6), NE); label("U", (-2, 6), NW); label("P", (0.5, 4), E); label("B", (0, 0), S); [/asy]$

Let $AB$ be the height. Since medians divide each other into a $2:1$ ratio, and the medians have length 12, we have $PC=MP=8$ and $UP=VP=4$. From right triangle $\triangle{MUP}$, $$MU^2=MP^2+UP^2=8^2+4^2=80,$$ so $MU=\sqrt{80}=4\sqrt{5}$. Since $CU$ is a median, $AM=8\sqrt{5}$. From right triangle $\triangle{MPC}$, $$MC^2=MP^2+PC^2=8^2+8^2=128,$$ which implies $MC=\sqrt{128}=8\sqrt{2}$. By symmetry $MB=\dfrac{8\sqrt{2}}{2}=4\sqrt{2}$.

Applying the Pythagorean Theorem to right triangle $\triangle{MAB}$ gives $AB^2=AM^2-MB^2=8\sqrt{5}^2-4\sqrt{2}^2=288$, so $AB=\sqrt{288}=12\sqrt{2}$. Then the area of $\triangle{AMC}$ is $$\dfrac{AB \cdot MC}{2}=\dfrac{8\sqrt{2} \cdot 12\sqrt{2}}{2}=\dfrac{96 \cdot 2}{2}=\boxed{\textbf{(C) }96}$$

## Solution 6 (Drawing)

By similarity, the area of $AUV$ is equal to $\frac{1}{4}$.

The area of $UVCM$ is equal to 72.

Assuming the total area of the triangle is S, the equation will be : $\frac{3}{4}$S = 72.

S = $\boxed{96}$

## Solution 7

Given a triangle with perpendicular medians with lengths $x$ and $y$, the area will be $\frac{2xy}{3}=\boxed{\textbf{(C) }96}$.

## Solution 8 (Fastest)

Connect the line segment $UV$ and it's easy to see quadrilateral $UVMC$ has an area of the product of its diagonals divided by $2$ which is $72$. Now, solving for triangle $AUV$ could be an option, but the drawing shows the area of $AUV$ will be less than the quadrilateral meaning the the area of $AMC$ is less than $72*2$ but greater than $72$, leaving only one possible answer choice, $\boxed{\textbf{(C) } 96}$.

-Rohan S.

## Solution 9

$[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((0,12)--(0,0)); label("M", (-4,0), W); label("C", (4,0), E); label("A", (0, 12), N); label("V", (2, 6), NE); label("U", (-2, 6), NW); label("P", (0.5, 4), E); label("B", (0, 0), S); [/asy]$

Connect $AP$, and let $B$ be the point where $AP$ intersects $MC$. $MB=CB$ because all medians of a triangle intersect at one point, which in this case is $P$. $MP:PV=2:1$ because the point at which all medians intersect divides the medians into segments of ratio $2:1$, so $MP=8$ and similarly $CP=8$. We apply the Pythagorean Theorem to triangle $MPC$ and get $MC=\sqrt{128}=8\sqrt{2}$. The area of triangle $MPC$ is $\dfrac{MP\cdot CP}{2}=32$, and that must equal to $\dfrac{MC\cdot BP}{2}$, so $BP=\dfrac{8}{\sqrt{2}}=4\cdot\sqrt{2}$. $BP=\dfrac{1}{3}BA$, so $BA=12\sqrt{2}$. The area of triangle $AMC$ is equal to $\dfrac{MC\cdot BA}{2}=\dfrac{8 \cdot \sqrt{2} \cdot 12 \cdot \sqrt{2}}{2}=\boxed{\textbf{(C)}\ 96}$.

-SmileKat32

## Solution 10 (High IQ)

$[\square MUVC] = 72$. Let intersection of line $AP$ and base $MC$ be $B$ $$[AUV]=[MUB]=[UVB]=[BVC] \implies \left[\frac{\triangle AMC}{4}\right] = \left[\frac{\square MUVC}{3}\right] \implies [AMC] = \boxed{\textbf{(C)}\ 96}$$

~herobrine-india

## Solution 11 (Kite)

$[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((-2,6)--(2,6)); label("M", (-4,0), W); label("C", (4,0), E); label("A", (0, 12), N); label("V", (2, 6), NE); label("U", (-2, 6), NW); label("P", (0, 3.6), S); [/asy]$

Since $\overline{MV}$ and $\overline{CU}$ intersect at a right angle, this means $MUVC$ is a kite. Hence, the area of this kite is $\frac{12 \cdot 12}{2} = 72$.

Also, notice that $\triangle AUV \sim \triangle AMC$ by AAA Similarity. Since the ratios of its sides is $\frac{1}{2}$, the ratios of the area is $\left(\frac{1}{2}\right)^2=\frac{1}{4}$. Therefore,

$$[AMC] = [MUVC] + \frac{1}{4} \cdot [AMC]$$

Simplifying gives us $\frac{3}{4} \cdot [AMC] = 72$, so $[AMC] = 72 \cdot \frac{4}{3} = \boxed{\textbf{(C)}\ 96}$

~MrThinker

## Solution 12 (Educated guessing)

Horizontally translate line $\overline{UC}$ until point $U$ is at point $V$, with $C$ subsequently at $C'$, and then connect up $C'$ and $C$ to create $\triangle MVC'$, which is a right triangle.

$[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((-2,6)--(2,6)); draw((2,6)--(8,0)); draw((4,0)--(8,0)); label("M", (-4,0), W); label("C", (4,0), S); label("A", (0, 12), N); label("V", (2, 6), NE); label("U", (-2, 6), NW); label("P", (0, 3.6), S); label("C'", (8,0), E); [/asy]$

Notice that $\triangle MVC'$ = $12 \cdot 12 \cdot \frac{1}{2} = 72$, and $\triangle MVC'$ = $\triangle MVC + \triangle MUV$ (since the latter has the same base and height as the sub-triangle $\triangle CVC'$ inside $\triangle MVC'$).

From this, we can deduce that $\boxed{\textbf{(B)}\ 72}$ cannot be true, since an incomplete part of $\triangle AMC$ is equal to it. We can also deduce that $\boxed{\textbf{(D)}\ 144}$ also cannot be true, since the unknown triangle $\triangle AUV = \triangle MUV$, and $\triangle MUV = \triangle CVC' < \triangle MVC'$. Therefore, the answer must be between $72$ and $144$, leaving $\boxed{\textbf{(C)}\ 96}$ as the only possible correct answer.

~DifrractSliver

## Video Solution 1

Education, The Study of Everything

~IceMatrix

~savannahsolver

~ pi_is_3.14