Difference between revisions of "2020 AMC 10A Problems/Problem 12"
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− | == 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> | ||
− | + | <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> | ||
<math>\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192</math> | <math>\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192</math> | ||
− | == Solution == | + | == Solution 1 == |
Since quadrilateral <math>UVCM</math> has perpendicular diagonals, its area can be found as half of the product of the length of the diagonals. Also note that <math>\triangle AUV</math> has <math>\frac 14</math> the area of triangle <math>AMC</math> by similarity, so <math>[UVCM]=\frac 34\cdot [AMC].</math> Thus, | Since quadrilateral <math>UVCM</math> has perpendicular diagonals, its area can be found as half of the product of the length of the diagonals. Also note that <math>\triangle AUV</math> has <math>\frac 14</math> the area of triangle <math>AMC</math> by similarity, so <math>[UVCM]=\frac 34\cdot [AMC].</math> Thus, | ||
<cmath>\frac 12 \cdot 12\cdot 12=\frac 34 \cdot [AMC]</cmath> | <cmath>\frac 12 \cdot 12\cdot 12=\frac 34 \cdot [AMC]</cmath> | ||
Line 11: | Line 22: | ||
<cmath>[AMC]=96\rightarrow \boxed{\textbf{(C)}}.</cmath> | <cmath>[AMC]=96\rightarrow \boxed{\textbf{(C)}}.</cmath> | ||
− | + | ==Solution 2 (Trapezoid)== | |
− | ==Solution 2 | ||
− | |||
− | |||
− | |||
<asy> | <asy> | ||
draw((-4,0)--(4,0)--(0,12)--cycle); | draw((-4,0)--(4,0)--(0,12)--cycle); | ||
Line 29: | 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 44: | Line 51: | ||
This is <math>\frac{3}{4}</math> of the triangle, so the area of the triangle is <math>\frac{4}{3}\cdot{72}=\boxed{\textbf{(C) } 96}</math> ~quacker88, diagram by programjames1 | This is <math>\frac{3}{4}</math> of the triangle, so the area of the triangle is <math>\frac{4}{3}\cdot{72}=\boxed{\textbf{(C) } 96}</math> ~quacker88, diagram by programjames1 | ||
− | ==Solution 4== | + | ==Solution 3 (Medians)== |
− | ( | + | Draw median <math>\overline{AB}</math>. |
+ | <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 <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>. | ||
+ | |||
+ | 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 (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 <math>AU = UM</math>, <math>AV = VC</math>, so <math>UV = \frac{1}{2} MC</math>. | ||
+ | |||
+ | As <math>\angle UPM = \angle VPC = 90</math>, we can see that <math>\triangle UPM \cong \triangle VPC</math> and <math>\triangle UVP \sim \triangle MPC</math> with a side ratio of <math>1 : 2</math>. | ||
− | <math> | + | So <math>UP = VP = 4</math>, <math>MP = PC = 8</math>. |
− | + | With that, we can see that <math>[\triangle UPM] = 16</math>, and the area of trapezoid <math>MUVC</math> is 72. | |
− | ==Solution 5 ( | + | As said in solution 1, <math>[\triangle AMC] = 72 / \frac{3}{4} = \boxed{\textbf{(C) } 96}</math>. |
− | + | ||
+ | -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 <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> | ||
+ | |||
+ | ==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. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | -Rohan S. | ||
+ | |||
+ | ==Solution 9== | ||
<asy> | <asy> | ||
draw((-4,0)--(4,0)--(0,12)--cycle); | draw((-4,0)--(4,0)--(0,12)--cycle); | ||
Line 67: | Line 152: | ||
</asy> | </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== | ||
− | + | Education, The Study of Everything | |
− | + | https://youtu.be/0TslJ3aDXac | |
− | ==Video Solution== | + | ==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== |
Latest revision as of 23:14, 1 May 2021
Contents
- 1 Problem 12
- 2 Solution 1
- 3 Solution 2 (Trapezoid)
- 4 Solution 3 (Medians)
- 5 Solution 4 (Triangles)
- 6 Solution 5 (Only Pythagorean Theorem)
- 7 Solution 6 (Drawing)
- 8 Solution 7
- 9 Solution 8 (Fastest)
- 10 Solution 9
- 11 Solution 10 (High IQ)
- 12 Video Solution 1
- 13 Video Solution 2
- 14 Video Solution 3
- 15 Video Solution 4
- 16 See Also
Problem 12
Triangle is isosceles with . Medians and are perpendicular to each other, and . What is the area of
Solution 1
Since quadrilateral has perpendicular diagonals, its area can be found as half of the product of the length of the diagonals. Also note that has the area of triangle by similarity, so Thus,
Solution 2 (Trapezoid)
We know that , and since the ratios of its sides are , the ratio of of their areas is .
If is the area of , then trapezoid is the area of .
Let's call the intersection of and . Let . Then . Since , and are heights of triangles and , respectively. Both of these triangles have base .
Area of
Area of
Adding these two gives us the area of trapezoid , which is .
This is of the triangle, so the area of the triangle is ~quacker88, diagram by programjames1
Solution 3 (Medians)
Draw median .
Since we know that all medians of a triangle intersect at the centroid, we know that passes through point . We also know that medians of a triangle divide each other into segments of ratio . Knowing this, we can see that , and since the two segments sum to , and are and , respectively.
Finally knowing that the medians divide the triangle into sections of equal area, finding the area of is enough. .
The area of . Multiplying this by gives us
~quacker88
Solution 4 (Triangles)
We know that , , so .
As , we can see that and with a side ratio of .
So , .
With that, we can see that , and the area of trapezoid is 72.
As said in solution 1, .
-QuadraticFunctions, solution 1 by ???
Solution 5 (Only Pythagorean Theorem)
Let be the height. Since medians divide each other into a ratio, and the medians have length 12, we have and . From right triangle , so . Since is a median, . From right triangle , which implies . By symmetry .
Applying the Pythagorean Theorem to right triangle gives , so . Then the area of is
Solution 6 (Drawing)
By similarity, the area of is equal to .
The area of is equal to 72.
Assuming the total area of the triangle is S, the equation will be : S = 72.
S =
Solution 7
Given a triangle with perpendicular medians with lengths and , the area will be .
Solution 8 (Fastest)
Connect the line segment and it's easy to see quadrilateral has an area of the product of its diagonals divided by which is . Now, solving for triangle could be an option, but the drawing shows the area of will be less than the quadrilateral meaning the the area of is less than but greater than , leaving only one possible answer choice, .
-Rohan S.
Solution 9
Connect , and let be the point where intersects . because all medians of a triangle intersect at one point, which in this case is . because the point at which all medians intersect divides the medians into segments of ratio , so and similarly . We apply the Pythagorean Theorem to triangle and get . The area of triangle is , and that must equal to , so . , so . The area of triangle is equal to .
-SmileKat32
Solution 10 (High IQ)
. Let intersection of line and base be
~herobrine-india
Video Solution 1
Education, The Study of Everything
Video Solution 2
~IceMatrix
Video Solution 3
~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.