Difference between revisions of "2017 AMC 12B Problems/Problem 24"
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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: | ||
− | + | ||
− | + | \begin{align*} | |
− | + | [BEC] &=[CED]=[BEA]=\frac{x^3}{2(x^2+1)} \\ | |
− | + | [ABCD] &=[AED]+[DEC]+[CEB]+[BEA] \\ | |
− | + | \frac{(BC)(AB+CD)}{2} &= 17*[CEB]+ [CEB] + [CEB] + [CEB] \\ | |
− | + | \frac{x^3+x}{2} &= \frac{20x^3}{2(x^2+1)} \\ | |
− | + | \frac{x}{x^2+1} &= \frac{20x^3}{x^2+1} \\ | |
+ | (x^2+1)^2 &=20x^2 \\ | ||
+ | x^4-18x^2+1 &=0 \implies x^2=9+4\sqrt{5}=4+2(2\sqrt{5})+5 \\ | ||
+ | \end{align*} | ||
+ | |||
Therefore, the answer is <math>\boxed{\textbf{(D) } 2+\sqrt{5}}</math> | Therefore, the answer is <math>\boxed{\textbf{(D) } 2+\sqrt{5}}</math> | ||
− | |||
==Solution 2== | ==Solution 2== | ||
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Note that <math>E</math> cannot be the intersection of <math>AC</math> and <math>BD</math>, as that would mean <math>[AED] = [CEB]</math> | Note that <math>E</math> cannot be the intersection of <math>AC</math> and <math>BD</math>, as that would mean <math>[AED] = [CEB]</math> | ||
− | <cmath>\because \triangle BCD \sim \triangle ABC, \quad \therefore \frac{CD}{BC} = \frac{BC}{AB}, \quad CD = | + | <cmath>\because \triangle BCD \sim \triangle ABC, \quad \therefore \frac{CD}{BC} = \frac{BC}{AB}, \quad CD = BC \cdot \frac{BC}{AB} = b \cdot \frac{b}{a} = \frac{b^2}{a}</cmath> |
− | <cmath>[CEB] = ( \frac{BC}{ | + | <cmath>[CEB] = ( \frac{BC}{AC} )^2 \cdot [ABC] = ( \frac{b}{ \sqrt{a^2+b^2} } )^2 \cdot \frac{ab}{2} = \frac{ab^3}{2(a^2+b^2)}</cmath> |
− | <cmath>BF = \frac{ 2[CEB] }{BC} = \frac{ \frac{ab^3}{a^2+b^2} }{b} = \frac{ab^2}{a^2+b^2}</cmath> | + | <cmath>BF = \frac{ 2[CEB] }{BC} = \frac{ 2 \cdot \frac{ab^3}{ 2(a^2+b^2) } }{b} = \frac{ab^2}{a^2+b^2}</cmath> |
<cmath>\because \triangle BFE \sim \triangle ABC, \quad \therefore \frac{EF}{BF} = \frac{BC}{AB}, \quad EF = BF \cdot \frac{BC}{AB} = \frac{ab^2}{a^2+b^2} \cdot \frac{b}{a} = \frac{b^3}{a^2+b^2}</cmath> | <cmath>\because \triangle BFE \sim \triangle ABC, \quad \therefore \frac{EF}{BF} = \frac{BC}{AB}, \quad EF = BF \cdot \frac{BC}{AB} = \frac{ab^2}{a^2+b^2} \cdot \frac{b}{a} = \frac{b^3}{a^2+b^2}</cmath> | ||
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(a^2 + b^2)^2 - 3a^2b^2 }{ a^2b^2 } = \frac{ a^4 - a^2b^2 + b^4 }{ a^2b^2 } = 17</cmath> | (a^2 + b^2)^2 - 3a^2b^2 }{ a^2b^2 } = \frac{ a^4 - a^2b^2 + b^4 }{ a^2b^2 } = 17</cmath> | ||
− | <cmath>a^4 - a^2b^2 + b^4 = 17 a^2b^2, \quad a^4 + b^4 = 18 a^2b^2, \quad \frac{a^2}{b^2} + \frac{b^2}{a^2} 18</cmath> | + | <cmath>a^4 - a^2b^2 + b^4 = 17 a^2b^2, \quad a^4 + b^4 = 18 a^2b^2, \quad \frac{a^2}{b^2} + \frac{b^2}{a^2} = 18</cmath> |
Let <math>x = \frac{a}{b}</math>, | Let <math>x = \frac{a}{b}</math>, | ||
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<cmath>x^2 + \frac{1}{x^2} = 18, \quad x^4 - 18x^2 + 1 = 0, \quad x^2 = \frac{18 + \sqrt{324-4} }{2} = 9+ 4\sqrt{5}</cmath> | <cmath>x^2 + \frac{1}{x^2} = 18, \quad x^4 - 18x^2 + 1 = 0, \quad x^2 = \frac{18 + \sqrt{324-4} }{2} = 9+ 4\sqrt{5}</cmath> | ||
− | < | + | <cmath>\frac{a}{b} = \sqrt{ 9+ 4\sqrt{5} } = \sqrt{ 4+ 4\sqrt{5}+5 } = \boxed{\textbf{(D) } 2+ \sqrt{5}}</cmath> |
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
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~Zeric | ~Zeric | ||
+ | |||
+ | ==Solution 6== | ||
+ | |||
+ | This solution involves proving <math>\triangle AED \sim \triangle CEB</math>. | ||
+ | |||
+ | Let <math>E'</math> be the intersection of <math>AC</math> and <math>BD</math>. Label points <math>F</math> and <math>G</math> the same way as <math>\textbf{Solution 3}</math>. | ||
+ | |||
+ | |||
+ | <math>\angle AE'D = \angle CE'B = \frac{\pi}{2} = \angle AFE</math>. Additionally, <math>\frac{E'D}{FE} = \frac{E'D}{E'G} = \frac{AE'}{AF}</math>, so <math>\triangle AFE \sim \triangle AE'D</math> by SAS. Therefore, <math>\angle BAC = \angle FAE + \angle EAE' = \angle E'AD + \angle EAE' = \angle EAD</math>. | ||
+ | |||
+ | |||
+ | Next, <math>\angle AFE = \frac{\pi}{2} = \angle EGD</math> because <math>FG \parallel BC</math>. Also, <math>\pi = \angle FEA + \angle AED + \angle DEG = \angle FEA + \angle AED + \angle EAF = \frac{\pi}{2} + \angle AED</math>, so <math>\angle AED = \frac{\pi}{2}</math>. Therefore, <math>\triangle AED \sim \triangle ABC</math> by AA. Since <math>\triangle CEB \sim \triangle ABC</math>, <math>\triangle AED \sim \triangle CEB</math>. | ||
+ | |||
+ | |||
+ | Given <math>\frac{[AED]}{[CEB]} = 17</math>, we deduce that the ratio of corresponding side lengths of <math>AED</math> to <math>CEB</math> must be <math>\sqrt{17}</math>. Now, we set <math>BC = 1</math>, <math>AB = x</math>, and <math>CD = \frac{1}{x}</math>. Using the Pythagorean Theorem, <math>AD = \sqrt{\Big(x-\frac{1}{x}\Big)^2 + 1^2}</math>. Thus, <math>\sqrt{17} = \frac{AD}{CB} = \frac{\sqrt{\Big(x-\frac{1}{x}\Big)^2 + 1^2}}{1}</math>. Solving gives <math>x = 2+\sqrt{5}</math>. | ||
+ | |||
+ | |||
+ | Finally, <math>\frac{AB}{BC} = \frac{2+\sqrt{5}}{1} = \boxed{\textbf{(D) } 2+\sqrt{5}}</math>. | ||
+ | |||
+ | ~Zhixing | ||
==Video Solution by MOP 2024== | ==Video Solution by MOP 2024== |
Latest revision as of 18:06, 31 July 2024
Contents
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:
\begin{align*} [BEC] &=[CED]=[BEA]=\frac{x^3}{2(x^2+1)} \\ [ABCD] &=[AED]+[DEC]+[CEB]+[BEA] \\ \frac{(BC)(AB+CD)}{2} &= 17*[CEB]+ [CEB] + [CEB] + [CEB] \\ \frac{x^3+x}{2} &= \frac{20x^3}{2(x^2+1)} \\ \frac{x}{x^2+1} &= \frac{20x^3}{x^2+1} \\ (x^2+1)^2 &=20x^2 \\ x^4-18x^2+1 &=0 \implies x^2=9+4\sqrt{5}=4+2(2\sqrt{5})+5 \\ \end{align*}
Therefore, the answer is
Solution 2
Draw line through
, with
on
and
on
,
. WLOG let
,
,
. By weighted average
.
Meanwhile, . This follows from comparing the ratios of triangle DEG to CFE and triangle AEG to FEB, both pairs in which the two triangles share a height perpendicular to FG, and have base ratio
.
. We obtain
,
namely
.
The rest is the same as Solution 1.
Solution 3
Let ,
,
Note that cannot be the intersection of
and
, as that would mean
Let ,
Solution 4
Let . Then from the similar triangles condition, we compute
and
. Hence, the
-coordinate of
is just
. Since
lies on the unit circle, we can compute the
coordinate as
. By Shoelace, we want
Factoring out denominators and expanding by minors, this is equivalent to
This factors as
, so
and so the answer is
.
Solution 5
Let where
. Because
. Notice that the diagonals are perpendicular with slopes of
and
. Let the intersection of
and
be
, then
. However, because
is a trapezoid,
and
share the same area, therefore
is the reflection of
over the perpendicular bisector of
, which is
. We use the linear equations of the diagonals,
, to find the coordinates of
.
The y-coordinate of
is simply
The area of
is
. We apply shoelace theorem to solve for the area of
. The coordinates of the triangle are
, so the area is
Finally, we use the property that the ratio of areas equals
~Zeric
Solution 6
This solution involves proving .
Let be the intersection of
and
. Label points
and
the same way as
.
. Additionally,
, so
by SAS. Therefore,
.
Next, because
. Also,
, so
. Therefore,
by AA. Since
,
.
Given , we deduce that the ratio of corresponding side lengths of
to
must be
. Now, we set
,
, and
. Using the Pythagorean Theorem,
. Thus,
. Solving gives
.
Finally, .
~Zhixing
Video Solution by MOP 2024
~r00tsOfUnity
Notes
1) is the most relevant answer choice because it shares numbers with the givens of the problem.
2) It's a very good guess to replace finding the area of triangle AED with the area of the triangle DAF, where F is the projection of D onto AB(then find the closest answer choice).
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.