Difference between revisions of "2021 AIME I Problems/Problem 11"
MRENTHUSIASM (talk | contribs) m (→Solution 1) |
MRENTHUSIASM (talk | contribs) (We have an overlapping "area method". So, I replaced the original Sol 1 with Sol 4, and credited Leonard my dude who first came up with this solution. In this way, the page looks cleaner. Please PM me if you are displeased with this change.) |
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~MRENTHUSIASM (by Geometry Expressions) | ~MRENTHUSIASM (by Geometry Expressions) | ||
− | ==Solution 1 | + | ==Solution 1 (Cyclic Quadrilaterals, Similar Triangles, and Trigonometry)== |
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This solution refers to the <b>Diagram</b> section. | This solution refers to the <b>Diagram</b> section. | ||
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Two solutions follow from here: | Two solutions follow from here: | ||
− | ===Solution | + | ===Solution 1.1 (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*}{12} | <cmath>\begin{alignat*}{12} | ||
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~MRENTHUSIASM (inspired by Math Jams's <b>2021 AIME I Discussion</b>) | ~MRENTHUSIASM (inspired by Math Jams's <b>2021 AIME I Discussion</b>) | ||
− | ===Solution | + | ===Solution 1.2 (Area Formulas)=== |
Let the brackets denote areas. | Let the brackets denote areas. | ||
We find <math>[ABCD]</math> in two different ways: | We find <math>[ABCD]</math> in two different ways: | ||
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Finally, substituting this result into <math>(\bigstar)</math> gives <math>22\cos\theta=\frac{242}{59},</math> from which the answer is <math>242+59=\boxed{301}.</math> | Finally, substituting this result into <math>(\bigstar)</math> gives <math>22\cos\theta=\frac{242}{59},</math> from which the answer is <math>242+59=\boxed{301}.</math> | ||
− | ~MRENTHUSIASM | + | ~MRENTHUSIASM (credit given to Leonard my dude) |
===Remark (Ptolemy's Theorem)=== | ===Remark (Ptolemy's Theorem)=== | ||
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\end{align*}</cmath> | \end{align*}</cmath> | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
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+ | ==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). | ||
+ | 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> | ||
+ | That is, <math>\tan^2{\frac{\theta}{2}}=\frac{1-\cos^2{\frac{\theta}{2}}}{\cos^2{\frac{\theta}{2}}}=\frac{24}{35}</math> or <math>\frac{35}{24}</math> | ||
+ | Thus, <math>\cos^2{\frac{\theta}{2}}=\frac{35}{59}</math> or <math>\frac{24}{59}</math> | ||
+ | <cmath>\cos{\theta}=2\cos^2{\frac{\theta}{2}}-1=\frac{\pm11}{59}</cmath> | ||
+ | In this context, <math>\cos{\theta}>0</math>. Thus, <math>\cos{\theta}=\frac{11}{59}</math> | ||
+ | <cmath>Ans=22*\cos{\theta}=22*\frac{11}{59}=\frac{242}{59}=\frac{m}{n}</cmath> | ||
+ | <cmath>m+n=242+59=\boxed{301}</cmath> | ||
+ | ~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 <math>\sqrt{4 \cdot 6 + 5 \cdot 7} = \sqrt{59}.</math> [I don't believe this is correct... are the two diagonals of <math>ABCD</math> necessarily congruent? -peace09] WLOG we focus on diagonal <math>BD.</math> To find the diagonal of the inner quadrilateral, we drop the altitude from <math>A</math> and <math>C</math> and calculate the length of <math>A_1C_1.</math> Let <math>x</math> be <math>A_1D</math> (Thus <math>A_1B = \sqrt{59} - x.</math> By Pythagorean theorem, we have <cmath>49 - x^2 = 16 - (\sqrt{59} - x)^2 \implies 92 = 2\sqrt{59}x \implies x = \frac{46}{\sqrt{59}} = \frac{46\sqrt{59}}{59}.</cmath> Now let <math>y</math> be <math>C_1D.</math> (thus making <math>C_1B = \sqrt{59} - y</math>). Similarly, we have <cmath>36 - y^2 = 25 - (\sqrt{59} - y)^2 \implies 70 = 2\sqrt{59}y \implies y = \frac{35}{\sqrt{59}} = \frac{35\sqrt{59}}{59}.</cmath> We see that <math>A_1C_1</math>, the scaled down diagonal is just <math>x - y = \frac{11\sqrt{59}}{59},</math> which is <math>\frac{\frac{11\sqrt{59}}{59}}{\sqrt{59}} = \frac{11}{59}</math> times our original diagonal <math>BD,</math> implying a scale factor of <math>\frac{11}{59}.</math> Thus, due to perimeters scaling linearly, the perimeter of the new quadrilateral is simply <math>\frac{11}{59} \cdot 22 = \frac{242}{59},</math> making our answer <math>242+59 = \boxed{301}.</math> | ||
+ | |||
+ | -fidgetboss_4000 | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2021|n=I|num-b=10|num-a=12}} | {{AIME box|year=2021|n=I|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:50, 30 August 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 (Cyclic Quadrilaterals, Similar Triangles, and Trigonometry)
This solution refers to the Diagram section.
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.
Suppose and
intersect at
and let
It follows that
and
We obtain the following diagram:
In every cyclic quadrilateral, each pair of opposite angles is supplementary. So, we have (both supplementary to
) and
(both supplementary to
), from which
by AA, with the ratio of similitude
Similarly, we have
(both supplementary to
) and
(both supplementary to
), from which
by AA, with the ratio of similitude
We apply the Transitive Property to
and
- We get
so
by SAS, with the ratio of similitude
- We get
so
by SAS, with the ratio of similitude
From and
the perimeter of
is
Two solutions follow from here:
Solution 1.1 (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 1.2 (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
so
Since
we have
It follows that
Finally, substituting this result into
gives
from which the answer is
~MRENTHUSIASM (credit given to Leonard my dude)
Remark (Ptolemy's Theorem)
In we have
~MRENTHUSIASM
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
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.