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]=(x^3)/(2(x^2+1)) \\ | |
− | + | [ABCD] &=[AED]+[DEC]+[CEB]+[BEA] \\ | |
− | + | (AB+CD)(BC)/2 &= 17*[CEB]+ [CEB] + [CEB] + [CEB] \\ | |
− | + | (x^3+x)/2 &=(20x^3)/(2(x^2+1)) \\ | |
− | + | (x)(x^2+1) &=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 \\ | ||
+ | |||
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== |
Revision as of 02:51, 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]=(x^3)/(2(x^2+1)) \\ [ABCD] &=[AED]+[DEC]+[CEB]+[BEA] \\ (AB+CD)(BC)/2 &= 17*[CEB]+ [CEB] + [CEB] + [CEB] \\ (x^3+x)/2 &=(20x^3)/(2(x^2+1)) \\ (x)(x^2+1) &=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 \\
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
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.