Difference between revisions of "2017 AMC 12B Problems/Problem 24"
(→See Also) |
m (→Solution 6) |
||
(32 intermediate revisions by 15 users not shown) | |||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | Quadrilateral <math>ABCD</math> has right angles at <math>B</math> and <math>C</math>, | + | 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>\ | ||
− | + | <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 by | + | ==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: | ||
+ | |||
+ | \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> | ||
+ | |||
+ | ==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>. 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 <math>EG:FE</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. | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | [[File:2017AMC12BP24.png|300px|center]] | ||
+ | |||
+ | Let <math>AB = a</math>, <math>BC = b</math>, <math>AC = \sqrt{a^2+b^2}</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 = BC \cdot \frac{BC}{AB} = b \cdot \frac{b}{a} = \frac{b^2}{a}</cmath> | ||
+ | |||
+ | <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{ 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>EG = FG - EF = b - \frac{b^3}{a^2+b^2} = \frac{a^2b}{a^2+b^2}</cmath> | ||
+ | |||
+ | <cmath>[ABCD] = \frac{AB + CD}{2} \cdot BC = \frac{a + \frac{b^2}{a} }{2} \cdot b = \frac{ b(a^2+b^2) }{ 2a }</cmath> | ||
+ | |||
+ | <cmath>[ABE] = \frac12 \cdot AB \cdot EF = \frac12 \cdot a \cdot \frac{b^3}{a^2+b^2} = \frac{ab^3}{ 2(a^2+b^2) }</cmath> | ||
+ | |||
+ | <cmath>[CDE] = \frac12 \cdot CD \cdot EG = \frac12 \cdot \frac{a^2b}{a^2+b^2} \cdot \frac{a^2b}{a^2+b^2} = \frac{ab^3}{ 2(a^2+b^2) }</cmath> | ||
+ | |||
+ | <cmath>ADE = [ABCD] - [ABE] - [CEB] - [CDE] = \frac{ b(a^2+b^2) }{ 2a } - \frac{ab^3}{ 2(a^2+b^2) } - \frac{ab^3}{ 2(a^2+b^2) }- \frac{ab^3}{ 2(a^2+b^2) } = \frac{ b(a^2+b^2)^2 - 3a^2b^3 }{ 2a(a^2+b^2) } </cmath> | ||
+ | |||
+ | <cmath>\frac{ [ADE] }{ [CEB] } = \frac { \frac{ b(a^2+b^2)^2 - 3a^2b^3 }{ 2a(a^2+b^2) } }{ \frac{ab^3}{2(a^2+b^2)} } = \frac{ | ||
+ | (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> | ||
+ | |||
+ | Let <math>x = \frac{a}{b}</math>, | ||
+ | |||
+ | <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] | ||
+ | |||
+ | == Solution 4== | ||
+ | Let <math>A=(-1,4a), B=(-1,0), C=(1,0), D=\bigg(1,\frac{1}{a}\bigg)</math>. Then from the similar triangles condition, we compute <math>CE=\frac{4a}{\sqrt{4a^2+1}}</math> and <math>BE=\frac{2}{\sqrt{4a^2+1}}</math>. Hence, the <math>y</math>-coordinate of <math>E</math> is just <math>\frac{BE\cdot CE}{BC}=\frac{4a}{4a^2+1}</math>. Since <math>E</math> lies on the unit circle, we can compute the <math>x</math> coordinate as <math>\frac{1-4a^2}{4a^2+1}</math>. By Shoelace, we want <cmath>\frac{1}{2}\det\begin{bmatrix} | ||
+ | -1 & 4a & 1\\ | ||
+ | \frac{1-4a^2}{4a^2+1} & \frac{4a}{4a^2+1} & 1\\ | ||
+ | 1 & \frac{1}{a} & 1 | ||
+ | \end{bmatrix}=17\cdot\frac{1}{2}\cdot 2 \cdot \frac{4a}{4a^2+1}</cmath>Factoring out denominators and expanding by minors, this is equivalent to | ||
+ | <cmath>\frac{32a^4-8a^2+2}{2a(4a^2+1)}=\frac{68a}{4a^2+1} \Longrightarrow 16a^4-72a^2+1=0</cmath>This factors as <math>(4a^2-8a-1)(4a^2+8a-1)=0</math>, so <math>a=1+\frac{\sqrt{5}}{2}</math> and so the answer is <math> \textbf{(D) \ }</math>. | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | Let <math>C = (0,0), D=(\frac1a, 0), B = (0,1), A = (a,1)</math> where <math>a>1</math>. Because <math>BC = 1, a = \frac{AB}{BC}</math>. Notice that the diagonals are perpendicular with slopes of <math>\frac1a</math> and <math>-a</math>. Let the intersection of <math>AC</math> and <math>BD</math> be <math>F</math>, then <math>\triangle BFC \sim \triangle ABC</math>. However, because <math>ABCD</math> is a trapezoid, <math>\triangle</math><math>BCF</math> and <math>\triangle ADF</math> share the same area, therefore <math>\triangle</math><math>BCE</math> is the reflection of <math>\triangle</math><math>BCF</math> over the perpendicular bisector of <math>BC</math>, which is <math>y=\frac12</math>. We use the linear equations of the diagonals, <math>y = -ax + 1, y = \frac1a x</math>, to find the coordinates of <math>F</math>. <cmath>-ax+1 = \frac1ax \Longrightarrow x = \frac{1}{a+\frac1a} = \frac{a}{a^2+1}</cmath> <cmath>y = \frac1ax = \frac{1}{a^2+1}</cmath> | ||
+ | The y-coordinate of <math>E</math> is simply <math>1-\frac{1}{a^2+1} = \frac{a^2}{a^2+1}</math> | ||
+ | The area of <math>\triangle BCE</math> is <math>\frac12 \frac{a}{a^2+1}</math>. We apply shoelace theorem to solve for the area of <math>\triangle ADE</math>. The coordinates of the triangle are <math>\{(\frac{a}{a^2+1}, \frac{a^2}{a^2+1}), (a,1), (\frac1a, 0)\}</math>, so the area is | ||
+ | |||
+ | <cmath>\frac12 |\frac{a^3}{a^2+1} + \frac1a - \frac{a}{a^2+1} - \frac{a}{a^2+1}| = \frac12 |\frac{a^3-2a}{a^2+1} + \frac1a|</cmath> | ||
+ | <cmath>= \frac12 |\frac{a^4-2a^2}{a(a^2+1)} + \frac{a^2+1}{a(a^2+1)}| = \frac12 \frac{a^4-a^2+1}{a(a^2+1)}</cmath> | ||
+ | Finally, we use the property that the ratio of areas equals <math>17</math> <cmath>\frac{\frac12}{\frac12} \frac{\frac{a^4-a^2+1}{a(a^2+1)}}{\frac{a}{a^2+1}} = 17 \Rightarrow \frac{a^4-a^2+1}{a^2} = 17 \Rightarrow a^4-18a^2+1 = 0</cmath> | ||
+ | <cmath>a^2 = 9+4\sqrt{5} = (2+\sqrt{5})^2 \Longrightarrow a = \boxed{\textbf{(D) } 2+\sqrt{5}}</cmath> | ||
+ | |||
+ | ~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== | ||
+ | https://youtu.be/h92s2BxlohI | ||
+ | |||
+ | ~r00tsOfUnity | ||
+ | |||
+ | == Notes== | ||
+ | 1) <math>\sqrt{17}</math> 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== | ==See Also== | ||
− | {{AMC12 box|year=2017|ab=B|num-b=23| | + | {{AMC12 box|year=2017|ab=B|num-b=23|num-a=25}} |
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
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.