Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 8-13.
Visit Beast Academy ‚
Books for Ages 8-13 Beast Academy Online
AoPS Academy
Nationwide learning centers
for students in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Problem)
m (Solutions should be first-come-first serve.)
(38 intermediate revisions by 17 users not shown)
Line 1: Line 1:
−
== Problem ==
+
== Problem 12 ==  
−
 
+
Triangle <math>AMC</math> is isosceles with <math>AM = AC</math>. Medians <math>\overline{MV}</math> and <math>\overline{CU}</math> are perpendicular to each other, and <math>MV=CU=12</math>. What is the area of <math>\triangle AMC?</math>
−
Triangle <math>AMC</math> is isoceles with <math>AM = AC</math>. Medians <math>\overline{MV}</math> and <math>\overline{CU}</math> are perpendicular to each other, and <math>MV=CU=12</math>. What is the area of <math>\triangle AMC?</math>
 
  
 
<asy>
 
<asy>
Line 37: Line 36:
 
</asy>
 
</asy>
  
−
We know that <math>\triangle AUV \sim \triangle AMC</math>, and since the ratios of its sides are <math>\frac{1}{2}</math>, the ratio of of their areas is <math>(\frac{1}{2})^2=\frac{1}{4}</math>.  
+
We know that <math>\triangle AUV \sim \triangle AMC</math>, and since the ratios of its sides are <math>\frac{1}{2}</math>, the ratio of of their areas is <math>\left(\frac{1}{2}\right)^2=\frac{1}{4}</math>.  
  
 
If <math>\triangle AUV</math> is <math>\frac{1}{4}</math> the area of <math>\triangle AMC</math>, then trapezoid <math>MUVC</math> is <math>\frac{3}{4}</math> the area of <math>\triangle AMC</math>.  
 
If <math>\triangle AUV</math> is <math>\frac{1}{4}</math> the area of <math>\triangle AMC</math>, then trapezoid <math>MUVC</math> is <math>\frac{3}{4}</math> the area of <math>\triangle AMC</math>.  
Line 68: Line 67:
 
</asy>
 
</asy>
  
−
Since we know that all medians of a triangle intersect at the incenter, we know that <math>\overline{AB}</math> passes through point <math>P</math>. We also know that medians of a triangle divide each other into segments of ratio <math>2:1</math>. Knowing this, we can see that <math>\overline{PC}:\overline{UP}=2:1</math>, and since the two segments sum to <math>12</math>, <math>\overline{PC}</math> and <math>\overline{UP}</math> are <math>8</math> and <math>4</math>, respectively.
+
Since we know that all medians of a triangle intersect at the centroid, we know that <math>\overline{AB}</math> passes through point <math>P</math>. We also know that medians of a triangle divide each other into segments of ratio <math>2:1</math>. Knowing this, we can see that <math>\overline{PC}:\overline{UP}=2:1</math>, and since the two segments sum to <math>12</math>, <math>\overline{PC}</math> and <math>\overline{UP}</math> are <math>8</math> and <math>4</math>, respectively.
  
 
Finally knowing that the medians divide the triangle into <math>6</math> sections of equal area, finding the area of <math>\triangle PUM</math> is enough. <math>\overline{PC} = \overline{MP} = 8</math>.
 
Finally knowing that the medians divide the triangle into <math>6</math> sections of equal area, finding the area of <math>\triangle PUM</math> is enough. <math>\overline{PC} = \overline{MP} = 8</math>.
Line 95: Line 94:
 
So <math>UP = VP = 4</math>, <math>MP = PC = 8</math>.
 
So <math>UP = VP = 4</math>, <math>MP = PC = 8</math>.
  
−
With that, we can see that <math>S\triangle UPM = 16</math>, and the area of trapezoid <math>MUVC</math> is 72.
+
With that, we can see that <math>[\triangle UPM] = 16</math>, and the area of trapezoid <math>MUVC</math> is 72.
  
−
As said in solution 1, <math>S\triangle AMC = 72  /  \frac{3}{4} = \boxed{\textbf{(C) } 96}</math>.
+
As said in solution 1, <math>[\triangle AMC] = 72  /  \frac{3}{4} = \boxed{\textbf{(C) } 96}</math>.
  
 
-QuadraticFunctions, solution 1 by ???
 
-QuadraticFunctions, solution 1 by ???
  
−
==Solution 5 (Bashy)==
+
==Solution 5 (Only Pythagorean Theorem)==
 
<asy>
 
<asy>
 
draw((-4,0)--(4,0)--(0,12)--cycle);
 
draw((-4,0)--(4,0)--(0,12)--cycle);
Line 107: Line 106:
 
draw((2,6)--(-4,0));
 
draw((2,6)--(-4,0));
 
draw((0,12)--(0,0));
 
draw((0,12)--(0,0));
−
draw((-2,6)--(2,6));
 
−
label("K", (0, 6), NE);
 
 
label("M", (-4,0), W);
 
label("M", (-4,0), W);
 
label("C", (4,0), E);
 
label("C", (4,0), E);
Line 119: Line 116:
 
</asy>
 
</asy>
  
−
It is well known that medians divide each other into segments of <math>2:1</math> ratio. From this, we have <math>PC=MP=8</math> and <math>UP=UV=4</math>. From right triangle <math>\triangle{MPC}</math>, <math>MC^2=MP^2+MC^2=8^2+8^2=128</math>, which implies <math>MC=\sqrt{128}=8\sqrt{2}</math>. Then the area of <math>\triangle{AMC}</math> is <math>\dfrac{8\sqrt{2} \cdot AB}{2}</math>, so our goal is to find <math>AB</math>.
+
Let <math>AB</math> be the height. Since medians divide each other into a <math>2:1</math> ratio, and the medians have length 12, we have <math>PC=MP=8</math> and <math>UP=VP=4</math>. From right triangle <math>\triangle{MUP}</math>, <cmath>MU^2=MP^2+UP^2=8^2+4^2=80,</cmath> so <math>MU=\sqrt{80}=4\sqrt{5}</math>. Since <math>CU</math> is a median, <math>AM=8\sqrt{5}</math>. From right triangle <math>\triangle{MPC}</math>, <cmath>MC^2=MP^2+PC^2=8^2+8^2=128,</cmath> which implies <math>MC=\sqrt{128}=8\sqrt{2}</math>. By symmetry <math>MB=\dfrac{8\sqrt{2}}{2}=4\sqrt{2}</math>.  
  
 +
Applying the Pythagorean Theorem to right triangle <math>\triangle{MAB}</math> gives <math>AB^2=AM^2-MB^2=8\sqrt{5}^2-4\sqrt{2}^2=288</math>, so <math>AB=\sqrt{288}=12\sqrt{2}</math>. Then the area of <math>\triangle{AMC}</math> is <cmath>\dfrac{AB \cdot MC}{2}=\dfrac{8\sqrt{2} \cdot 12\sqrt{2}}{2}=\dfrac{96 \cdot 2}{2}=\boxed{\textbf{(C) }96}</cmath>
  
−
Note that <math>AB=AP+BP</math>. Since <math>\triangle{AMC}</math> is isosceles, by symmetry <math>MB=MC</math>, because <math>AB</math> is the altitude. Knowing this, <math>BP</math> is the median to hypotenuse <math>MC</math> of triangle <math>\triangle{MPC}</math>, which means <math>2BP=MC</math>. Since <math>MC=8\sqrt{2}</math>, <math>BP=4\sqrt{2}</math>.  
+
==Solution 6 (Drawing)==
 +
By similarity, the area of <math>AUV</math> is equal to <math>\frac{1}{4}</math>.
  
 +
The area of <math>UVCM</math> is equal to 72.
  
−
Now we find <math>AP</math>. Note <math>AP=AK+KP</math>. (K is the intersection of the diagonals of quadrilateral <math>AUPV</math>). From right triangle <math>\triangle{AVP}</math>, we have <math>UV=4\sqrt{2}</math> by the Pythagorean Theorem. By symmetry, <math>PK</math> is the median to hypotenuse <math>UV</math>, which means <math>2PK=UV</math>. This trivially means <math>PK=2\sqrt{2}</math>.  
+
Assuming the total area of the triangle is S, the equation will be : <math>\frac{3}{4}</math>S = 72.
  
 +
S = <math>\boxed{96}</math>
  
−
Notice quadrilateral <math>AUPV</math> is a kite, which means <math>\triangle{AKU}</math> is right(the diagonals are perpendicular). By the Pythagorean Theorem, <math>AK^2=AU^2-UK^2</math>. Since <math>CU</math> is a median, <math>AU=UM</math>. From right triangle <math>\triangle{UPM}</math>, <math>UM^2=UP^2+MP^2=4^2+8^2=80</math>, which means <math>UM=4\sqrt{5}</math>, and thus <math>AU=4\sqrt{5}</math>.  
+
==Solution 7==
 +
Given a triangle with perpendicular medians with lengths <math>x</math> and <math>y</math>, the area will be <math>\frac{2xy}{3}=\boxed{\textbf{(C) }96}</math>.
  
 +
==Solution 8 (Fastest)==
 +
Connect the line segment <math>UV</math> and it's easy to see quadrilateral <math>UVMC</math> has an area of the product of its diagonals divided by <math>2</math> which is <math>72</math>. Now, solving for triangle <math>AUV</math> could be an option, but the drawing shows the area of <math>AUV</math> will be less than the quadrilateral meaning the the area of <math>AMC</math> is less than <math>72*2</math> but greater than <math>72</math>, leaving only one possible answer choice, <math>\boxed{\textbf{(C) } 96}</math>.
  
−
From our previous equation <math>AK^2=AU^2-UK^2</math>, we thus have <cmath>AK^2=80-UK^2=80-\left(\dfrac{UV}{2}\right)=80-8=72,</cmath> so <math>AK=6\sqrt{2}</math>. We also know <math>PK=2\sqrt{2}</math>, so <math>AP=AK+PK=8\sqrt{2}</math>.  
+
-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 <math>AP</math>, and let <math>B</math> be the point where <math>AP</math> intersects <math>MC</math>. <math>MB=CB</math> because all medians of a triangle intersect at one point, which in this case is <math>P</math>. <math>MP:PV=2:1</math> because the point at which all medians intersect divides the medians into segments of ratio <math>2:1</math>, so <math>MP=8</math> and similarly <math>CP=8</math>. We apply the Pythagorean Theorem to triangle <math>MPC</math> and get <math>MC=\sqrt{128}=8\sqrt{2}</math>. The area of triangle <math>MPC</math> is <math>\dfrac{MP\cdot CP}{2}=32</math>, and that must equal to <math>\dfrac{MC\cdot BP}{2}</math>, so <math>BP=\dfrac{8}{\sqrt{2}}=4\cdot\sqrt{2}</math>. <math>BP=\dfrac{1}{3}BA</math>, so <math>BA=12\sqrt{2}</math>. The area of triangle <math>AMC</math> is equal to <math>\dfrac{MC\cdot BA}{2}=\dfrac{8 \cdot \sqrt{2} \cdot 12 \cdot \sqrt{2}}{2}=\boxed{\textbf{(C)} 96}</math>.
 +
 +
-SmileKat32
 +
 +
== Solution 10 (High IQ) ==
 +
<math>[\square MUVC] = 72</math>. Let intersection of line <math>AP</math> and base <math>MC</math> be <math>B</math> <cmath> [AUV]=[MUB]=[UVB]=[BVC] \implies \left[\frac{\triangle AMC}{4}\right] = \left[\frac{\square MUVC}{3}\right] \implies [AMC] = 96</cmath>
 +
 +
~herobrine-india 
 +
 +
==Video Solution 1==
  
−
Recall that <math>AB=AP+BP=8\sqrt{2}+4\sqrt{2}=12\sqrt{2}</math>. By the area formula, <cmath>[ABC]=\dfrac{AB \cdot MC}{2}=\dfrac{8\sqrt{2} \cdot 12\sqrt{2}}{2}=\dfrac{96 \cdot 2}{2}=\boxed{\textbf{(C) }96}.</cmath>
+
Education, The Study of Everything
  
−
==Video Solution==
+
https://youtu.be/0TslJ3aDXac
 +
 
 +
==Video Solution 2==
 
https://youtu.be/ZGwAasE32Y4
 
https://youtu.be/ZGwAasE32Y4
  
 
~IceMatrix
 
~IceMatrix
 +
 +
==Video Solution 3==
 +
https://youtu.be/7ZvKOYuwSnE
 +
 +
~savannahsolver
 +
 +
== Video Solution 4 ==
 +
https://youtu.be/4_x1sgcQCp4?t=2067
 +
 +
~ pi_is_3.14
  
 
==See Also==
 
==See Also==

Revision as of 23:14, 1 May 2021

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

-QuadraticFunctions, solution 1 by ???

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 $B$ \[[AUV]=[MUB]=[UVB]=[BVC] \implies \left[\frac{\triangle AMC}{4}\right] = \left[\frac{\square MUVC}{3}\right] \implies [AMC] = 96\]

~herobrine-india

Video Solution 1

Education, The Study of Everything

https://youtu.be/0TslJ3aDXac

Video Solution 2

https://youtu.be/ZGwAasE32Y4

~IceMatrix

Video Solution 3

https://youtu.be/7ZvKOYuwSnE

~savannahsolver

Video Solution 4

https://youtu.be/4_x1sgcQCp4?t=2067

~ pi_is_3.14

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_12&oldid=153086"
Invalid username
Login to AoPS