Difference between revisions of "2020 AMC 10A Problems/Problem 12"
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label("P", (0, 3.6), S); | label("P", (0, 3.6), S); | ||
</asy> | </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>. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | As said in solution 1, <math>S\triangle AMC = 72 / \frac{3}{4} = \boxed{\textbf{(C) } 96}</math>. | ||
+ | |||
+ | -QuadraticFunctions, solution 1 by ??? | ||
+ | |||
+ | ==Solution 5 (Triangles)== | ||
+ | [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] | ||
We know that <math>AU = UM</math>, <math>AV = VC</math>, so <math>UV = \frac{1}{2} MC</math>. | We know that <math>AU = UM</math>, <math>AV = VC</math>, so <math>UV = \frac{1}{2} MC</math>. | ||
Revision as of 12:57, 1 February 2020
Contents
Problem
Triangle is isoceles 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 incenter, 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 (Triangles)
[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] 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 ???
Video Solution
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.