Difference between revisions of "2017 AMC 12B Problems/Problem 24"
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Quadrilateral <math>ABCD</math> has right angles at <math>B</math> and <math>C</math>, Triangle <math>ABC</math> ~ Triangle <math>BCD</math>, and <math>AB > BC</math>. There is a point <math>E</math> in the interior of <math>ABCD</math> such that Triangle <math>ABC</math> ~ Triangle <math>CEB</math> and the area of Triangle <math>AED</math> is <math>17</math> times the area of Triangle <math>CEB</math>. What is <math>AB/BC</math> | Quadrilateral <math>ABCD</math> has right angles at <math>B</math> and <math>C</math>, Triangle <math>ABC</math> ~ Triangle <math>BCD</math>, and <math>AB > BC</math>. There is a point <math>E</math> in the interior of <math>ABCD</math> such that Triangle <math>ABC</math> ~ Triangle <math>CEB</math> and the area of Triangle <math>AED</math> is <math>17</math> times the area of Triangle <math>CEB</math>. What is <math>AB/BC</math> | ||
− | <math>\textbf{(A) } 1+\sqrt | + | <math>\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } 2 + \sqrt{2} \qquad \textbf{(C) } \sqrt{17} \qquad \textbf{(D) } 2 + \sqrt{5} \qquad \textbf{(E) } 1 + 2\sqrt{3}</math> |
==Solution== | ==Solution== |
Revision as of 22:24, 16 February 2017
Problem
Quadrilateral has right angles at
and
, Triangle
~ Triangle
, and
. There is a point
in the interior of
such that Triangle
~ Triangle
and the area of Triangle
is
times the area of Triangle
. What is
Solution
Solution by TorrTar
Let ,
,
. Note that
. The Pythagorean theorem states that
. Since
, the ratios of side lengths must be equal. Since
,
and
. Let Point F be a point on
such that
is an altitude of triangle
. Note that
, so
and
can be calculated. Solving for these lengths gives
and
. Since
and
form altitudes of triangles
and
, respectively, the areas of these triangles can be calculated. Additionally, the area of triangle
can be calculated, as it is a right triangle. Solving for each of these yields:
(Minus yields a negative value)
Thus the answer is D: 2+sqrt(5)
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.