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

(Solution 3 (Trapezoid))
(Solution 4)
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==Solution 4==
 
==Solution 4==
Again, call the intersection of the medians <math>P</math>. There are many ways to find the lengths of
+
(using Solution 3 diagram)
  
<math>\triangle UPV \sim \triangle CPM</math>, and since <math>\frac{\overline{UV}}{\overline{MC}}=\frac{1}{2}</math>, <math>\frac{\overline{UP}}{\overline{PC}}=\frac{1}{2}</math>, and since <math>\overline{UP} + \overline{PC} = 12</math>, <math>\overline{UP}</math> and <math>\overline{PC}</math> are <math>4</math> and <math>8</math>, respectively. Proceed with the Pythagorean Theorem like the other solutions.
+
<math>\triangle UPV \sim \triangle CPM</math>, and <math>\frac{\overline{UV}}{\overline{MC}}=\frac{\overline{UP}}{\overline{PC}}=\frac{1}{2}</math>. Since <math>\overline{UP} + \overline{PC} = 12</math><math>\text{, }</math> <math>\text{ }</math> <math>\overline{UP}</math> and <math>\overline{PC}</math> are <math>4</math> and <math>8</math>, respectively.
 +
 
 +
By symmetry, <math>\overline{PC}=\overline{PM}=8</math>. We can now do Pythagorean Theorem on <math>\triangle UPM</math>.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 23:36, 31 January 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

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 (CHEATING)

Draw a to-scale diagram with your graph paper and straightedge. Measure the height and approximate the area.

Solution 3 (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 4

(using Solution 3 diagram)

$\triangle UPV \sim \triangle CPM$, and $\frac{\overline{UV}}{\overline{MC}}=\frac{\overline{UP}}{\overline{PC}}=\frac{1}{2}$. Since $\overline{UP} + \overline{PC} = 12$$\text{, }$ $\text{ }$ $\overline{UP}$ and $\overline{PC}$ are $4$ and $8$, respectively.

By symmetry, $\overline{PC}=\overline{PM}=8$. We can now do Pythagorean Theorem on $\triangle UPM$.

Video Solution

https://youtu.be/ZGwAasE32Y4

~IceMatrix

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AMC 10 Problems and Solutions

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