Difference between revisions of "2021 AIME I Problems/Problem 11"
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− | ==Solution== | + | ==Solution 1== |
Let <math>O</math> be the intersection of <math>AC</math> and <math>BD</math>. Let <math>\theta = \angle AOB</math>. | Let <math>O</math> be the intersection of <math>AC</math> and <math>BD</math>. Let <math>\theta = \angle AOB</math>. | ||
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From Ptolemy's theorem, we have that <math>(BD)(AC) = 4\times6+5\times7 = 59</math>. From Brahmagupta's Formula, <math>[ABCD] = \sqrt{(11-4)(11-5)(11-6)(11-7)} = 2\sqrt{210}</math>. But the area is also <math>\frac{1}{2}(BD)(AC)\sin\theta = \frac{59}{2}\sin\theta</math>, so <math>\sin \theta = \frac{4\sqrt{210}}{59} \implies \cos \theta = \frac{11}{59}</math>. Then the desired fraction is <math>(4+5+6+7)\cos\theta = \frac{242}{59}</math> for an answer of <math>\boxed{301}</math>. | From Ptolemy's theorem, we have that <math>(BD)(AC) = 4\times6+5\times7 = 59</math>. From Brahmagupta's Formula, <math>[ABCD] = \sqrt{(11-4)(11-5)(11-6)(11-7)} = 2\sqrt{210}</math>. But the area is also <math>\frac{1}{2}(BD)(AC)\sin\theta = \frac{59}{2}\sin\theta</math>, so <math>\sin \theta = \frac{4\sqrt{210}}{59} \implies \cos \theta = \frac{11}{59}</math>. Then the desired fraction is <math>(4+5+6+7)\cos\theta = \frac{242}{59}</math> for an answer of <math>\boxed{301}</math>. | ||
− | ==Finding | + | ==Solution 2 (Finding cos{x})== |
The angle <math>\theta</math> between diagonals satisfies <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{(s-b)(s-d}{(s-a)(s-c)}}</cmath> (see https://en.wikipedia.org/wiki/Cyclic_quadrilateral#Angle_formulas). | The angle <math>\theta</math> between diagonals satisfies <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{(s-b)(s-d}{(s-a)(s-c)}}</cmath> (see https://en.wikipedia.org/wiki/Cyclic_quadrilateral#Angle_formulas). | ||
Thus, <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-4)(11-6)}{(11-5)(11-7)}}</cmath> or <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-5)(11-7)}{(11-4)(11-6)}}</cmath> | Thus, <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-4)(11-6)}{(11-5)(11-7)}}</cmath> or <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-5)(11-7)}{(11-4)(11-6)}}</cmath> |
Revision as of 12:58, 21 April 2021
Contents
Problem
Let be a cyclic quadrilateral with
and
. Let
and
be the feet of the perpendiculars from
and
, respectively, to line
and let
and
be the feet of the perpendiculars from
and
respectively, to line
. The perimeter of
is
, where
and
are relatively prime positive integers. Find
.
Diagram
Solution 1
Let be the intersection of
and
. Let
.
Firstly, since , we deduce that
is cyclic. This implies that
, with a ratio of
. This means that
. Similarly,
. Hence
It therefore only remains to find
.
From Ptolemy's theorem, we have that . From Brahmagupta's Formula,
. But the area is also
, so
. Then the desired fraction is
for an answer of
.
Solution 2 (Finding cos{x})
The angle between diagonals satisfies
(see https://en.wikipedia.org/wiki/Cyclic_quadrilateral#Angle_formulas).
Thus,
or
That is,
or
Thus,
or
In this context,
. Thus,
~y.grace.yu
Solution 3 (Pythagorean Theorem)
We assume that the two quadrilateral mentioned in the problem are similar (due to both of them being cyclic). Note that by Ptolemy’s, one of the diagonals has length WLOG we focus on diagonal
To find the diagonal of the inner quadrilateral, we drop the altitude from
and
and calculate the length of
Let
be
(Thus
By Pythagorean theorem, we have
Now let
be
(thus making
). Similarly, we have
We see that
, the scaled down diagonal is just
which is
times our original diagonal
implying a scale factor of
Thus, due to perimeters scaling linearly, the perimeter of the new quadrilateral is simply
making our answer
-fidgetboss_4000
See also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.