Difference between revisions of "2018 AIME I Problems/Problem 15"
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==Solution 4== | ==Solution 4== | ||
− | Let the sides of the quadrilaterals be <math>a,b,c,</math> and <math>d</math> in some order such that <math>A</math> has <math>a</math> opposite of <math>c</math>, <math>B</math> has <math>a</math> opposite of <math>b</math>, and <math>C</math> has <math>a</math> opposite of <math>d</math>. Then let the diagonals of <math>A</math> be <math>e</math> and <math>f</math>. Similarly to solution <math>2</math>, we get that <math>\tfrac{2}{3}(ac+bd)=\tfrac{3}{5}(ab+cd)=\tfrac{6}{7}(ad+bc)=2K</math>, but this is also equal to <math>2\cdot\tfrac{eab+ecd}{4(1)}=2\cdot\tfrac{fad+fbc}{4(1)}</math> using the area formula for a triangle using the circumradius and the sides, so <math>\tfrac{e(ab+cd)}{2}=\tfrac{3}{5}(ab+cd)</math> and <math>\tfrac{f(ad+bc)}{2}=\tfrac{6}{7}(ad+bc)</math>. Solving for <math>e</math> and <math>f</math>, we get that <math>e=\tfrac{6}{5}</math> and <math>f=\tfrac{12}{7}</math>, but <math>K=\tfrac{1}{2}\cdot\tfrac{2}{3}\cdot{}ef</math>, similarly to solution <math>2</math>, so <math>K=\tfrac{24}{35}</math> and the answer is <math>24+35=\boxed{059}</math>. | + | Let the sides of the quadrilaterals be <math>a,b,c,</math> and <math>d</math> in some order such that <math>A</math> has <math>a</math> opposite of <math>c</math>, <math>B</math> has <math>a</math> opposite of <math>b</math>, and <math>C</math> has <math>a</math> opposite of <math>d</math>. Then, let the diagonals of <math>A</math> be <math>e</math> and <math>f</math>. Similarly to solution <math>2</math>, we get that <math>\tfrac{2}{3}(ac+bd)=\tfrac{3}{5}(ab+cd)=\tfrac{6}{7}(ad+bc)=2K</math>, but this is also equal to <math>2\cdot\tfrac{eab+ecd}{4(1)}=2\cdot\tfrac{fad+fbc}{4(1)}</math> using the area formula for a triangle using the circumradius and the sides, so <math>\tfrac{e(ab+cd)}{2}=\tfrac{3}{5}(ab+cd)</math> and <math>\tfrac{f(ad+bc)}{2}=\tfrac{6}{7}(ad+bc)</math>. Solving for <math>e</math> and <math>f</math>, we get that <math>e=\tfrac{6}{5}</math> and <math>f=\tfrac{12}{7}</math>, but <math>K=\tfrac{1}{2}\cdot\tfrac{2}{3}\cdot{}ef</math>, similarly to solution <math>2</math>, so <math>K=\tfrac{24}{35}</math> and the answer is <math>24+35=\boxed{059}</math>. |
==See Also== | ==See Also== | ||
{{AIME box|year=2018|n=I|num-b=14|after=Last question}} | {{AIME box|year=2018|n=I|num-b=14|after=Last question}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 09:59, 4 August 2022
Contents
[hide]Problem 15
David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, , which can each be inscribed in a circle with radius
. Let
denote the measure of the acute angle made by the diagonals of quadrilateral
, and define
and
similarly. Suppose that
,
, and
. All three quadrilaterals have the same area
, which can be written in the form
, where
and
are relatively prime positive integers. Find
.
Solution 1
Suppose our four sides lengths cut out arc lengths of ,
,
, and
, where
. Then, we only have to consider which arc is opposite
. These are our three cases, so
Our first case involves quadrilateral
with
,
,
, and
.
Then, by Law of Sines, and
. Therefore,
so our answer is
.
Note that the conditions of the problem are satisfied when the lengths of the four sticks are about .
By S.B.
Solution 2
Let the four stick lengths be ,
,
, and
. WLOG, let’s say that quadrilateral
has sides
and
opposite each other, quadrilateral
has sides
and
opposite each other, and quadrilateral
has sides
and
opposite each other. The area of a convex quadrilateral can be written as
, where
and
are the lengths of the diagonals of the quadrilateral and
is the angle formed by the intersection of
and
. By Ptolemy's theorem
for quadrilateral
, so, defining
as the area of
,
Similarly, for quadrilaterals
and
,
and
Multiplying the three equations and rearranging, we see that
The circumradius
of a cyclic quadrilateral with side lengths
,
,
, and
and area
can be computed as
.
Inserting what we know,
So our answer is
.
~Solution by divij04
Solution 3 (No words)
Shelomovskii, vvsss, www.deoma-cmd.ru
Solution 4
Let the sides of the quadrilaterals be and
in some order such that
has
opposite of
,
has
opposite of
, and
has
opposite of
. Then, let the diagonals of
be
and
. Similarly to solution
, we get that
, but this is also equal to
using the area formula for a triangle using the circumradius and the sides, so
and
. Solving for
and
, we get that
and
, but
, similarly to solution
, so
and the answer is
.
See Also
2018 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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