Difference between revisions of "2017 AMC 12B Problems/Problem 24"
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==Problem== | ==Problem== | ||
− | Quadrilateral <math>ABCD</math> has right angles at <math>B</math> and <math>C</math>, <math>\triangle ABC | + | Quadrilateral <math>ABCD</math> has right angles at <math>B</math> and <math>C</math>, <math>\triangle ABC \sim \triangle BCD</math>, and <math>AB > BC</math>. There is a point <math>E</math> in the interior of <math>ABCD</math> such that <math>\triangle ABC \sim \triangle CEB</math> and the area of <math>\triangle AED</math> is <math>17</math> times the area of <math>\triangle CEB</math>. What is <math>\tfrac{AB}{BC}</math>? |
+ | |||
<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> | <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 1== | ||
Let <math>CD=1</math>, <math>BC=x</math>, and <math>AB=x^2</math>. Note that <math>AB/BC=x</math>. By the Pythagorean Theorem, <math>BD=\sqrt{x^2+1}</math>. Since <math>\triangle BCD \sim \triangle ABC \sim \triangle CEB</math>, the ratios of side lengths must be equal. Since <math>BC=x</math>, <math>CE=\frac{x^2}{\sqrt{x^2+1}}</math> and <math>BE=\frac{x}{\sqrt{x^2+1}}</math>. Let F be a point on <math>\overline{BC}</math> such that <math>\overline{EF}</math> is an altitude of triangle <math>CEB</math>. Note that <math>\triangle CEB \sim \triangle CFE \sim \triangle EFB</math>. Therefore, <math>BF=\frac{x}{x^2+1}</math> and <math>CF=\frac{x^3}{x^2+1}</math>. Since <math>\overline{CF}</math> and <math>\overline{BF}</math> form altitudes of triangles <math>CED</math> and <math>BEA</math>, respectively, the areas of these triangles can be calculated. Additionally, the area of triangle <math>BEC</math> can be calculated, as it is a right triangle. Solving for each of these yields: | Let <math>CD=1</math>, <math>BC=x</math>, and <math>AB=x^2</math>. Note that <math>AB/BC=x</math>. By the Pythagorean Theorem, <math>BD=\sqrt{x^2+1}</math>. Since <math>\triangle BCD \sim \triangle ABC \sim \triangle CEB</math>, the ratios of side lengths must be equal. Since <math>BC=x</math>, <math>CE=\frac{x^2}{\sqrt{x^2+1}}</math> and <math>BE=\frac{x}{\sqrt{x^2+1}}</math>. Let F be a point on <math>\overline{BC}</math> such that <math>\overline{EF}</math> is an altitude of triangle <math>CEB</math>. Note that <math>\triangle CEB \sim \triangle CFE \sim \triangle EFB</math>. Therefore, <math>BF=\frac{x}{x^2+1}</math> and <math>CF=\frac{x^3}{x^2+1}</math>. Since <math>\overline{CF}</math> and <math>\overline{BF}</math> form altitudes of triangles <math>CED</math> and <math>BEA</math>, respectively, the areas of these triangles can be calculated. Additionally, the area of triangle <math>BEC</math> can be calculated, as it is a right triangle. Solving for each of these yields: | ||
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− | + | ==Solution 2== | |
+ | Draw line <math>FG</math> through <math>E</math>, with <math>F</math> on <math>BC</math> and <math>G</math> on <math>AD</math>, <math>FG \parallel AB</math>. WLOG let <math>CD=1</math>, <math>CB=x</math>, <math>AB=x^2</math>. By weighted average <math>FG=\frac{1+x^4}{1+x^2}</math>. | ||
+ | |||
+ | Meanwhile, <math>FE:EG=[\triangle CBE]:[\triangle ADE]=1:17</math>. | ||
+ | |||
+ | <math>FE=\frac{x^2}{1+x^2}</math>. We obtain <math>\frac{1+x^4}{1+x^2}=\frac{18x^2}{1+x^2}</math>, | ||
+ | namely <math>x^4-18x^2+1=0</math>. | ||
− | + | The rest is the same as Solution 1. | |
− | |||
− | |||
− | |||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=B|num-b=23|num-a=25}} | {{AMC12 box|year=2017|ab=B|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:01, 19 February 2017
Contents
[hide]Problem
Quadrilateral has right angles at
and
,
, and
. There is a point
in the interior of
such that
and the area of
is
times the area of
. What is
?
Solution 1
Let ,
, and
. Note that
. By the Pythagorean Theorem,
. Since
, the ratios of side lengths must be equal. Since
,
and
. Let F be a point on
such that
is an altitude of triangle
. Note that
. Therefore,
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:
Therefore, the answer is
Solution 2
Draw line through
, with
on
and
on
,
. WLOG let
,
,
. By weighted average
.
Meanwhile, .
. We obtain
,
namely
.
The rest is the same as Solution 1.
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 |
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