Difference between revisions of "2020 AMC 10A Problems/Problem 12"

Revision as of 05:15, 1 February 2020

Problem

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

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

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 $(\frac{1}{2})^2=\frac{1}{4}$ .

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 incenter, 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 (Similar triangles)

Video Solution

https://youtu.be/ZGwAasE32Y4

~IceMatrix

@@ Line 61: / Line 61: @@
 The area of <math>\triangle PUM = \frac{4\cdot8}{2}=16</math>. Multiplying this by <math>6</math> gives us <math>6\cdot16=\boxed{\textbf{(C) }96}</math> ~quacker88
+==Solution 4 (Similar triangles)==
 ==Video Solution==

2020 AMC 10A (Problems • Answer Key • Resources)
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