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Difference between revisions of "2015 AMC 12A Problems/Problem 12"

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==Solution==
 
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Clearly, the parabolas must intersect the x-axis at the same two points. Their distance multiplied by <math>4 - (-2)</math> (the distance between the y-intercepts), all divided by 2 is equal to 12, the area of the kite (half the product of the diagonals). That distance is thus 4, and so the x-intercepts are <math>(2, 0), (-2, 0)</math>. Then <math>0 = 4a - 2 \rightarrow a = 0.5</math>, and <math>0 = 4 - 4b \rightarrow b = 1</math>. Then <math>a + b = 1.5</math>, or <math>\textbf{(C)}</math>.
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Clearly, the parabolas must intersect the x-axis at the same two points. Their distance multiplied by <math>4 - (-2)</math> (the distance between the y-intercepts), all divided by 2 is equal to 12, the area of the kite (half the product of the diagonals). That distance is thus 4, and so the x-intercepts are <math>(2, 0), (-2, 0).</math> Then <math>0 = 4a - 2 \rightarrow a = 0.5</math>, and <math>0 = 4 - 4b \rightarrow b = 1.</math> Then <math>a + b = 1.5</math>, or <math>\textbf{(C)}</math>.

Revision as of 23:41, 4 February 2015

Problem 12

The parabolas $y=ax^2 - 2$ and $y=4 - bx^2$ intersect the coordinate axes in exactly four points, and these four points are the vertices of a kite of area $12$. What is $a+b$?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 1.5\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}}\ 2.5\qquad\textbf{(E)}\ 3$ (Error compiling LaTeX. Unknown error_msg)


Solution

Clearly, the parabolas must intersect the x-axis at the same two points. Their distance multiplied by $4 - (-2)$ (the distance between the y-intercepts), all divided by 2 is equal to 12, the area of the kite (half the product of the diagonals). That distance is thus 4, and so the x-intercepts are $(2, 0), (-2, 0).$ Then $0 = 4a - 2 \rightarrow a = 0.5$, and $0 = 4 - 4b \rightarrow b = 1.$ Then $a + b = 1.5$, or $\textbf{(C)}$.

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