Difference between revisions of "2017 AMC 12B Problems/Problem 24"
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\begin{align*} | \begin{align*} | ||
− | [BEC] &=[CED]=[BEA]= | + | [BEC] &=[CED]=[BEA]=\frac{x^3}{2(x^2+1)} \\ |
[ABCD] &=[AED]+[DEC]+[CEB]+[BEA] \\ | [ABCD] &=[AED]+[DEC]+[CEB]+[BEA] \\ | ||
− | (AB+CD) | + | \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^2+1)^2 &=20x^2 \\ | ||
x^4-18x^2+1 &=0 \implies x^2=9+4\sqrt{5}=4+2(2\sqrt{5})+5 \\ | x^4-18x^2+1 &=0 \implies x^2=9+4\sqrt{5}=4+2(2\sqrt{5})+5 \\ |
Revision as of 02:54, 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
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.