Difference between revisions of "2021 AIME I Problems/Problem 11"
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===Solution 4.1 (Law of Cosines)=== | ===Solution 4.1 (Law of Cosines)=== | ||
Note that <math>\cos(180^\circ-\theta)=-\cos\theta</math> holds for all <math>\theta.</math> We apply the Law of Cosines to <math>\triangle ABE, \triangle BCE, \triangle CDE,</math> and <math>\triangle DAE,</math> respectively: | Note that <math>\cos(180^\circ-\theta)=-\cos\theta</math> holds for all <math>\theta.</math> We apply the Law of Cosines to <math>\triangle ABE, \triangle BCE, \triangle CDE,</math> and <math>\triangle DAE,</math> respectively: | ||
− | <cmath>\begin{alignat*}{ | + | <cmath>\begin{alignat*}{12} |
− | AB^2& | + | &&&AE^2+BE^2-2\cdot AE\cdot BE\cdot\cos\theta&&=AB^2&&\quad\implies\quad AE^2+BE^2-2\cdot AE\cdot BE\cdot\cos\theta&&=4^2, \hspace{15mm} &(1\star) \\ |
− | BC^2& | + | &&&BE^2+CE^2-2\cdot BE\cdot CE\cdot\cos(180^\circ-\theta)&&=BC^2&&\quad\implies\quad BE^2+CE^2+2\cdot BE\cdot CE\cdot\cos\theta&&=5^2, \hspace{15mm} &(2\star) \\ |
− | CD^2& | + | &&&CE^2+DE^2-2\cdot CE\cdot DE\cdot\cos\theta&&=CD^2&&\quad\implies\quad CE^2+DE^2-2\cdot CE\cdot DE\cdot\cos\theta&&=6^2, \hspace{15mm} &(3\star) \\ |
− | DA^2& | + | &&&DE^2+AE^2-2\cdot DE\cdot AE\cdot\cos(180^\circ-\theta)&&=DA^2&&\quad\implies\quad DE^2+AE^2+2\cdot DE\cdot AE\cdot\cos\theta&&=7^2. \hspace{15mm} &(4\star) \\ |
\end{alignat*}</cmath> | \end{alignat*}</cmath> | ||
We subtract <math>(1\star)+(3\star)</math> from <math>(2\star)+(4\star):</math> | We subtract <math>(1\star)+(3\star)</math> from <math>(2\star)+(4\star):</math> | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
+ | 2\cdot AE\cdot BE\cdot\cos\theta+2\cdot BE\cdot CE\cdot\cos\theta+2\cdot CE\cdot DE\cdot\cos\theta+2\cdot DE\cdot AE\cdot\cos\theta&=22 \\ | ||
2\cdot\cos\theta\cdot(\phantom{ }\underbrace{AE\cdot BE+BE\cdot CE+CE\cdot DE+DE\cdot AE}_{\text{Use the result from }\textbf{Remark}\text{.}}\phantom{ })&=22 \\ | 2\cdot\cos\theta\cdot(\phantom{ }\underbrace{AE\cdot BE+BE\cdot CE+CE\cdot DE+DE\cdot AE}_{\text{Use the result from }\textbf{Remark}\text{.}}\phantom{ })&=22 \\ | ||
2\cdot\cos\theta\cdot59&=22 \\ | 2\cdot\cos\theta\cdot59&=22 \\ |
Revision as of 05:31, 30 May 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
~MRENTHUSIASM (by Geometry Expressions)
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 [I don't believe this is correct... are the two diagonals of
necessarily congruent? -peace09] 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
Solution 4 (Cyclic Quadrilaterals, Similar Triangles, and Ptolemy's Theorem)
This solution refers to the Diagram section.
Suppose and
intersect at
and let
By the Converse of the Inscribed Angle Theorem, if distinct points and
lie on the same side of
(but not on
itself) for which
then
and
are cyclic. From the Converse of the Inscribed Angle Theorem, quadrilaterals
and
are all cyclic.
We obtain the following diagram:
In every cyclic quadrilateral, each pair of opposite angles is supplementary. So, we have and
by angle chasing, from which
by AA, with the ratio of similitude
Similarly, we have
and
by angle chasing, from which
by AA, with the ratio of similitude
We apply the Transitive Property to
and
- We get
from which
by SAS, with the ratio of similitude
- We get
from which
by SAS, with the ratio of similitude
From and
the perimeter of
is
Two solutions follow from here:
Solution 4.1 (Law of Cosines)
Note that holds for all
We apply the Law of Cosines to
and
respectively:
We subtract
from
Finally, substituting this result into
gives
from which the answer is
~MRENTHUSIASM (inspired by Math Jams's 2021 AIME I Discussion)
Solution 4.2 (Area Formulas)
Let the brackets denote areas.
We find in two different ways:
- Note that
holds for all
By area addition, we get
- By Brahmagupta's Formula, we get
where
is the semiperimeter of
Equating the expressions for we have
from which
Since
we know that
It follows that
Finally, substituting this result into
gives
from which the answer is
~MRENTHUSIASM
Remark (Ptolemy's Theorem)
In we have
~MRENTHUSIASM
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.