2015 AMC 12A Problems/Problem 12: Difference between revisions
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==Solution== | ==Solution== | ||
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 = | 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>. | ||
== 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}} | ||
Revision as of 00:04, 11 February 2017
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 
?
Solution
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 
.
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 | |
