Difference between revisions of "2017 AMC 12B Problems/Problem 24"
Rockmanex3 (talk | contribs) (→See Also) |
(→Solution 3) |
||
Line 34: | Line 34: | ||
\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 | \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>. | <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>. | ||
+ | |||
+ | == 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== |
Revision as of 00:16, 31 December 2019
Problem 24
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.
Solution 3
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
.
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.