Difference between revisions of "2014 AMC 10A Problems/Problem 14"
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(Used more elementary language in the solution and shrunk diagram.) |
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==Solution== | ==Solution== | ||
<asy>//Needs refining | <asy>//Needs refining | ||
− | size( | + | size(12cm); |
fill((0,10)--(6,8)--(0,-10)--cycle,rgb(.7,.7,.7)); | fill((0,10)--(6,8)--(0,-10)--cycle,rgb(.7,.7,.7)); | ||
for(int i=-2;i<=8;i+=1) | for(int i=-2;i<=8;i+=1) | ||
Line 23: | Line 23: | ||
draw(rightanglemark((0,10),(6,8),(0,-10),20)); | draw(rightanglemark((0,10),(6,8),(0,-10),20)); | ||
label("$A$",(6,8),NE); | label("$A$",(6,8),NE); | ||
+ | label("$a$", (0,5),W); | ||
+ | label("$a$",(0,-5),W); | ||
+ | label("$a$",(3,4),NW); | ||
+ | |||
</asy> | </asy> | ||
− | Note that the <math>y</math>- | + | Note that if the <math>y</math>-intercepts have a sum of <math>0</math>, the distance from the origin to each of the intercepts must be the same. Call this distance <math>a</math>. Since the <math>\angle PAQ = 90^\circ</math>, the length of the median to the midpoint of the hypotenuse is equal to half the length of the hypotenuse. Since the median's length is <math>\sqrt{6^2+8^2} = 10</math>, this means <math>a=10</math>, and the length of the hypotenuse is <math>2a = 20</math>. Since the <math>x</math>-coordinate of <math>A</math> is the same as the altitude to the hypotenuse, <math>[APQ] = \dfrac{20 \cdot 6}{2} = \boxed{\textbf{(D)} \: 60}</math>. |
==See Also== | ==See Also== |
Revision as of 23:20, 7 February 2014
Problem
The -intercepts, and , of two perpendicular lines intersecting at the point have a sum of zero. What is the area of ?
Solution
Note that if the -intercepts have a sum of , the distance from the origin to each of the intercepts must be the same. Call this distance . Since the , the length of the median to the midpoint of the hypotenuse is equal to half the length of the hypotenuse. Since the median's length is , this means , and the length of the hypotenuse is . Since the -coordinate of is the same as the altitude to the hypotenuse, .
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.