Difference between revisions of "2015 AMC 12A Problems/Problem 12"
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− | ==Problem== | + | == Problem == |
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The parabolas <math>y=ax^2 - 2</math> and <math>y=4 - bx^2</math> intersect the coordinate axes in exactly four points, and these four points are the vertices of a kite of area <math>12</math>. What is <math>a+b</math>? | The parabolas <math>y=ax^2 - 2</math> and <math>y=4 - bx^2</math> intersect the coordinate axes in exactly four points, and these four points are the vertices of a kite of area <math>12</math>. What is <math>a+b</math>? | ||
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 1.5\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 2.5\qquad\textbf{(E)}\ 3</math> | <math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 1.5\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 2.5\qquad\textbf{(E)}\ 3</math> | ||
+ | == Solutions == | ||
+ | === Solution 1 === | ||
+ | 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 = \boxed{\textbf{(B)}\ 1.5}</math>. | ||
− | ==Solution== | + | === Solution 2 === |
− | + | The parabolas must intersect the x-axis at the same two points for the kite to form. We find the x values at which they intersect by equating them and solving for x as shown below. <math>y = ax^2-2</math> and <math>y = 4 - bx^2\rightarrow ax^2-2 = 4-bx^2\rightarrow (a+b)x^2 = 6 \rightarrow x = +\sqrt{\dfrac{6}{a+b}}</math> or <math>-\sqrt{\dfrac{6}{a+b}}</math>. The x-values of the y-intercepts is 0, so we plug in zero in each of them and get <math>-2</math> and <math>4</math>. The area of a kite is <math>\dfrac{d_1*d_2}{2}</math>. The <math>d_1</math> is <math>2+4 = 6</math>. The <math>d_2</math> is <math>2\sqrt{\dfrac{6}{a+b}}</math>. Solving for the area <math>\rightarrow \dfrac{1}{2}*(6)*(2*\sqrt{\dfrac{6}{a+b}}) = 12 \rightarrow (2*\sqrt{\dfrac{6}{a+b}}) = 4 \rightarrow (\sqrt{\dfrac{6}{a+b}}) = 2 \rightarrow \dfrac{6}{a+b} = 4 \rightarrow | |
+ | \dfrac{6}{4} = (a+b)</math>, therefore <math>a + b = \boxed{\textbf{(B)}\ 1.5}</math>. | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2015|ab=A|num-b=11|num-a=13}} | {{AMC12 box|year=2015|ab=A|num-b=11|num-a=13}} |
Latest revision as of 14:18, 19 January 2021
Problem
The parabolas and
intersect the coordinate axes in exactly four points, and these four points are the vertices of a kite of area
. What is
?
Solutions
Solution 1
Clearly, the parabolas must intersect the x-axis at the same two points. Their distance multiplied by (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
Then
, and
Then
.
Solution 2
The parabolas must intersect the x-axis at the same two points for the kite to form. We find the x values at which they intersect by equating them and solving for x as shown below. and
or
. The x-values of the y-intercepts is 0, so we plug in zero in each of them and get
and
. The area of a kite is
. The
is
. The
is
. Solving for the area
, therefore
.
See Also
2015 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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 12 Problems and Solutions |