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Difference between revisions of "2014 AMC 10A Problems/Problem 14"

(Solution)
(Used more elementary language in the solution and shrunk diagram.)
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==Solution==
 
==Solution==
 
<asy>//Needs refining
 
<asy>//Needs refining
size(20cm);
+
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)
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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>-coordinates of <math>P</math> and <math>Q</math> must be negations of each other; hence it follows that <math>O=(0,0)</math> is the midpoint of <math>\overline{PQ}</math>. Because <math>\triangle APQ</math> is right, the median <math>\overline{OA}</math> is its circumradius and <math>O</math> its circumcentre. The [[Pythagorean Theorem]] gives us <math>AO=\sqrt{6^2+8^2}=10</math>, so <math>OP=OQ=10</math>. Thus <math>P</math> has <math>y</math>-coordinate 10 and <math>Q</math> <math>y</math>-coordinate -10 and <math>PQ=20</math>. The altitude from <math>A</math> to <math>\overline{PQ}</math> is <math>6</math> (<math>\overline{PQ}</math> is on the <math>y</math>-axis, hence this altitude is just the <math>x</math>-coordinate of <math>A</math>), so <math>[\triangle APQ]=\dfrac{20\cdot6}{2}=60\implies\boxed{\textbf{(D)}\ 60}</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 $y$-intercepts, $P$ and $Q$, of two perpendicular lines intersecting at the point $A(6,8)$ have a sum of zero. What is the area of $\triangle APQ$?

$\textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72$

Solution

[asy]//Needs refining size(12cm); fill((0,10)--(6,8)--(0,-10)--cycle,rgb(.7,.7,.7)); for(int i=-2;i<=8;i+=1)   draw((i,-12)--(i,12),grey); for(int j=-12;j<=12;j+=1)   draw((-2,j)--(8,j),grey); draw((-3,0)--(9,0),linewidth(1),Arrows); //x-axis draw((0,-13)--(0,13),linewidth(1),Arrows); //y-axis dot((0,0)); dot((6,8)); draw((-2,10.66667)--(8,7.33333),Arrows); draw((7.33333,12)--(-0.66667,-12),Arrows); draw((6,8)--(0,8)); draw((6,8)--(0,0)); draw(rightanglemark((0,10),(6,8),(0,-10),20)); label("$A$",(6,8),NE); label("$a$", (0,5),W); label("$a$",(0,-5),W); label("$a$",(3,4),NW);  [/asy] Note that if the $y$-intercepts have a sum of $0$, the distance from the origin to each of the intercepts must be the same. Call this distance $a$. Since the $\angle PAQ = 90^\circ$, 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 $\sqrt{6^2+8^2} = 10$, this means $a=10$, and the length of the hypotenuse is $2a = 20$. Since the $x$-coordinate of $A$ is the same as the altitude to the hypotenuse, $[APQ] = \dfrac{20 \cdot 6}{2} = \boxed{\textbf{(D)} \: 60}$.

See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
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

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