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

Revision as of 00:20, 8 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, $[APQ] = \dfrac{20 \cdot 6}{2} = \boxed{\textbf{(D)} \: 60}$ .

2014 AMC 10A (Problems • Answer Key • Resources)
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Difference between revisions of "2014 AMC 10A Problems/Problem 14"

Revision as of 00:20, 8 February 2014

Problem

Solution

See Also

@@ Line 7: / Line 7: @@
 ==Solution==
 <asy>//Needs refining
-size(20cm);
+size(12cm);
 fill((0,10)--(6,8)--(0,-10)--cycle,rgb(.7,.7,.7));
 for(int i=-2;i<=8;i+=1)
@@ Line 23: / Line 23: @@
 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 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==