Difference between revisions of "2021 AIME I Problems/Problem 11"
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MRENTHUSIASM (talk | contribs) (→Solution 4 (Cyclic Quadrilaterals, Similar Triangles, Law of Cosines, Ptolemy's Theorem)) |
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Suppose <math>\overline{AC}</math> and <math>\overline{BD}</math> intersect at <math>E,</math> and let <math>\theta=\angle AEB.</math> | Suppose <math>\overline{AC}</math> and <math>\overline{BD}</math> intersect at <math>E,</math> and let <math>\theta=\angle AEB.</math> | ||
− | By the <b>Converse of the Inscribed Angle Theorem</b>, if distinct points <math>X</math> and <math>Y</math> lie on the same side of <math>\overline{PQ}</math> (but not on <math>\overline{PQ}</math> itself) for which <math>\angle PXQ=\angle PYQ,</math> then <math>P,Q,X,</math> and <math>Y</math> are cyclic. | + | By the <b>Converse of the Inscribed Angle Theorem</b>, if distinct points <math>X</math> and <math>Y</math> lie on the same side of <math>\overline{PQ}</math> (but not on <math>\overline{PQ}</math> itself) for which <math>\angle PXQ=\angle PYQ,</math> then <math>P,Q,X,</math> and <math>Y</math> are cyclic. From the Converse of the Inscribed Angle Theorem, we conclude that quadrilaterals <math>ABA_1B_1,BCC_1B_1,CDC_1D_1,</math> and <math>DAA_1D_1</math> are all cyclic. |
+ | |||
+ | In every cyclic quadrilateral, each pair of opposite angles is supplementary. So, we have <math>\angle EA_1B_1=\angle EAB</math> and <math>\angle EB_1A_1=\angle EBA</math> by angle chasing, from which <math>\triangle A_1B_1E \sim \triangle ABE</math> by AA, with the ratio of similitude <cmath>\frac{A_1B_1}{AB}=\underbrace{\frac{A_1E}{AE}}_{\substack{\text{right} \\ \triangle A_1AE}}=\underbrace{\frac{B_1E}{BE}}_{\substack{\text{right} \\ \triangle B_1BE}}=\cos\theta.</cmath> | ||
<b>IN PROGRESS. NO EDIT PLEASE. A MILLION THANKS. | <b>IN PROGRESS. NO EDIT PLEASE. A MILLION THANKS. |
Revision as of 18:07, 29 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, Law of Cosines, Ptolemy's Theorem)
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, we conclude that quadrilaterals
and
are all cyclic.
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
IN PROGRESS. NO EDIT PLEASE. A MILLION THANKS.
I WILL FINISH WITHIN ONE DAY.
~MRENTHUSIASM (inspired by Math Jams's 2021 AIME I Discussion)
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 |
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