Difference between revisions of "2020 AMC 10A Problems/Problem 20"
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==Solution 3 (coordinates)== | ==Solution 3 (coordinates)== | ||
<asy> | <asy> | ||
+ | size(10cm,0); | ||
draw((10,30)--(10,0)--(-8,-6)--(-10,0)--(10,30)); | draw((10,30)--(10,0)--(-8,-6)--(-10,0)--(10,30)); | ||
+ | draw((-20,0)--(20,0)); | ||
+ | draw((0,-15)--(0,35)); | ||
+ | draw((10,30)--(-8,-6)); | ||
draw(circle((0,0),10)); | draw(circle((0,0),10)); | ||
− | label("D",(10,30), | + | label("E",(-4.05,-.25),S); |
− | label("C",(10,0), | + | label("D",(10,30),NE); |
− | label("B",(-8,-6), | + | label("C",(10,0),NE); |
− | label("A",(-10,0), | + | label("B",(-8,-6),SW); |
+ | label("A",(-10,0),NW); | ||
+ | label("5",(-10,0)--(-5,0), NE); | ||
+ | label("15",(-5,0)--(10,0), N); | ||
+ | label("30",(10,0)--(10,30), E); | ||
+ | dot((-5,0)); | ||
+ | dot((-10,0)); | ||
+ | dot((-8,-6)); | ||
+ | dot((10,0)); | ||
+ | dot((10,30)); | ||
</asy> | </asy> | ||
Let the points be <math>A(-10,0)</math>, <math>\:B(x,y)</math>, <math>\:C(10,0)</math>, <math>\:D(10,30)</math>,and <math>\:E(-5,0)</math>, respectively. Since <math>B</math> lies on line <math>DE</math>, we know that <math>y=2x+10</math>. Furthermore, since <math>\angle{ABC}=90^\circ</math>, <math>B</math> lies on the circle with diameter <math>AC</math>, so <math>x^2+y^2=100</math>. Solving for <math>x</math> and <math>y</math> with these equations, we get the solutions <math>(0,10)</math> and <math>(-8,-6)</math>. We immediately discard the <math>(0,10)</math> solution as <math>y</math> should be negative. Thus, we conclude that <math>[ABCD]=[ACD]+[ABC]=\frac{20\cdot30}{2}+\frac{20\cdot6}{2}=\boxed{\textbf{(D)}\:360}</math>. | Let the points be <math>A(-10,0)</math>, <math>\:B(x,y)</math>, <math>\:C(10,0)</math>, <math>\:D(10,30)</math>,and <math>\:E(-5,0)</math>, respectively. Since <math>B</math> lies on line <math>DE</math>, we know that <math>y=2x+10</math>. Furthermore, since <math>\angle{ABC}=90^\circ</math>, <math>B</math> lies on the circle with diameter <math>AC</math>, so <math>x^2+y^2=100</math>. Solving for <math>x</math> and <math>y</math> with these equations, we get the solutions <math>(0,10)</math> and <math>(-8,-6)</math>. We immediately discard the <math>(0,10)</math> solution as <math>y</math> should be negative. Thus, we conclude that <math>[ABCD]=[ACD]+[ABC]=\frac{20\cdot30}{2}+\frac{20\cdot6}{2}=\boxed{\textbf{(D)}\:360}</math>. |
Revision as of 22:37, 1 February 2020
Contents
Problem
Quadrilateral satisfies and Diagonals and intersect at point and What is the area of quadrilateral
Solution 1 (Just Drop An Altitude)
It's crucial to draw a good diagram for this one. Since and , we get . Now we need to find to get the area of the whole quadrilateral. Drop an altitude from to and call the point of intersection . Let . Since , then . By dropping this altitude, we can also see two similar triangles, and . Since is , and , we get that . Now, if we redraw another diagram just of , we get that . Now expanding, simplifying, and dividing by the GCF, we get . This factors to . Since lengths cannot be negative, . Since , . So
(I'm very sorry if you're a visual learner)
~Ultraman
Solution 2 (Pro Guessing Strats)
We know that the big triangle has area 300. Use the answer choices which would mean that the area of the little triangle is a multiple of 10. Thus the product of the legs is a multiple of 20. Guess that the legs are equal to and , and because the hypotenuse is 20 we get . Testing small numbers, we get that when and , is indeed a square. The area of the triangle is thus 60, so the answer is .
~tigershark22 ~(edited by HappyHuman)
Solution 3 (coordinates)
Let the points be , , , ,and , respectively. Since lies on line , we know that . Furthermore, since , lies on the circle with diameter , so . Solving for and with these equations, we get the solutions and . We immediately discard the solution as should be negative. Thus, we conclude that .
Solution 4 (Trigonometry)
Using the law of cosines, express and in terms of . The sum of these two equations is by the Pythagorean Theorem. Solving for , and using the fact that , we find . Since and , , which is five times ,
(This solution is incomplete, can someone complete it please-Lingjun) Latex edited by kc5170
Video Solution
~IceMatrix
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.