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

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-QuadraticFunctions, solution 1 by ???
 
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==Solution 5 (Pythagorean Theorem, Lots of symmetry)==
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==Solution 5 (Bashy)==
 
<asy>
 
<asy>
 
draw((-4,0)--(4,0)--(0,12)--cycle);
 
draw((-4,0)--(4,0)--(0,12)--cycle);

Revision as of 12:31, 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 (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 $S\triangle UPM = 16$, and the area of trapezoid $MUVC$ is 72.

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

-QuadraticFunctions, solution 1 by ???

Solution 5 (Bashy)

[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((0,12)--(0,0)); draw((-2,6)--(2,6)); label("K", (0, 6), NE); 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]

It is well known that medians divide each other into segments of $2:1$ ratio. From this, we have $PC=MP=8$ and $UP=UV=4$. From right triangle $\triangle{MPC}$, $MC^2=MP^2+MC^2=8^2+8^2=128$, which implies $MC=\sqrt{128}=8\sqrt{2}$. Then the area of $\triangle{AMC}$ is $\dfrac{8\sqrt{2} \cdot AB}{2}$, so our goal is to find $AB$.


Note that $AB=AP+BP$. Since $\triangle{AMC}$ is isosceles, by symmetry $MB=MC$, because $AB$ is the altitude. Knowing this, $BP$ is the median to hypotenuse $MC$ of triangle $\triangle{MPC}$, which means $2BP=MC$. Since $MC=8\sqrt{2}$, $BP=4\sqrt{2}$.


Now we find $AP$. Note $AP=AK+KP$. (K is the intersection of the diagonals of quadrilateral $AUPV$). From right triangle $\triangle{AVP}$, we have $UV=4\sqrt{2}$ by the Pythagorean Theorem. By symmetry, $PK$ is the median to hypotenuse $UV$, which means $2PK=UV$. This trivially means $PK=2\sqrt{2}$.


Notice quadrilateral $AUPV$ is a kite, which means $\triangle{AKU}$ is right(the diagonals are perpendicular). By the Pythagorean Theorem, $AK^2=AU^2-UK^2$. Since $CU$ is a median, $AU=UM$. From right triangle $\triangle{UPM}$, $UM^2=UP^2+MP^2=4^2+8^2=80$, which means $UM=4\sqrt{5}$, and thus $AU=4\sqrt{5}$.


From our previous equation $AK^2=AU^2-UK^2$, we thus have \[AK^2=80-UK^2=80-\left(\dfrac{UV}{2}\right)=80-8=72,\] so $AK=6\sqrt{2}$. We also know $PK=2\sqrt{2}$, so $AP=AK+PK=8\sqrt{2}$.


Recall that $AB=AP+BP=8\sqrt{2}+4\sqrt{2}=12\sqrt{2}$. By the area formula, \[[ABC]=\dfrac{AB \cdot MC}{2}=\dfrac{8\sqrt{2} \cdot 12\sqrt{2}}{2}=\dfrac{96 \cdot 2}{2}=\boxed{\textbf{(C) }96}.\]

Video Solution

https://youtu.be/ZGwAasE32Y4

~IceMatrix

See Also

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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

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