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

(See Also: problem no. incorrect)
(Solution)
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==Solution==
 
==Solution==
 +
<asy>//Needs refining
 +
size(20cm);
 +
fill((0,10)--(6,8)--(0,-10)--cycle,rgb(.7,.7,.7));
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for(int i=-2;i<=8;i+=1)
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  draw((i,-12)--(i,12),grey);
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for(int j=-12;j<=12;j+=1)
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  draw((-2,j)--(8,j),grey);
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draw((-3,0)--(9,0),linewidth(1),Arrows); //x-axis
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draw((0,-13)--(0,13),linewidth(1),Arrows); //y-axis
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dot((0,0));
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dot((6,8));
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draw((-2,10.66667)--(8,7.33333),Arrows);
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draw((7.33333,12)--(-0.66667,-12),Arrows);
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draw((6,8)--(0,8));
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draw((6,8)--(0,0));
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draw(rightanglemark((0,10),(6,8),(0,-10),20));
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label("$A$",(6,8),NE);
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</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>.
  
 
==See Also==
 
==See Also==

Revision as of 07:33, 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(20cm); 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); [/asy] Note that the $y$-coordinates of $P$ and $Q$ must be negations of each other; hence it follows that $O=(0,0)$ is the midpoint of $\overline{PQ}$. Because $\triangle APQ$ is right, the median $\overline{OA}$ is its circumradius and $O$ its circumcentre. The Pythagorean Theorem gives us $AO=\sqrt{6^2+8^2}=10$, so $OP=OQ=10$. Thus $P$ has $y$-coordinate 10 and $Q$ $y$-coordinate -10 and $PQ=20$. The altitude from $A$ to $\overline{PQ}$ is $6$ ($\overline{PQ}$ is on the $y$-axis, hence this altitude is just the $x$-coordinate of $A$), so $[\triangle APQ]=\dfrac{20\cdot6}{2}=60\implies\boxed{\textbf{(D)}\ 60}$.

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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