Difference between revisions of "2018 AIME I Problems/Problem 15"
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By S.B. | By S.B. | ||
+ | |||
+ | ===Note=== | ||
+ | [[File:CyclIntersect.png|400px]] | ||
+ | |||
+ | The solution uses <cmath>\varphi_A=a+c.</cmath> | ||
+ | |||
+ | We can see that this follows because <math>\varphi_A = \frac12 (2a+2c)=a+c,</math> where <math>a</math> and <math>c</math> are the central angles of opposite sides. | ||
==Solution 2== | ==Solution 2== | ||
Line 40: | Line 47: | ||
<cmath>2d=360^\circ-\varphi_A-\varphi_B-\varphi_C</cmath> | <cmath>2d=360^\circ-\varphi_A-\varphi_B-\varphi_C</cmath> | ||
− | + | Using the [[trigonometric identity|sum-to-product identities]], our area of the quadrilateral <math>K</math> then would be | |
− | + | <cmath> | |
− | + | \begin{align*} | |
− | + | K&=\frac{1}{2}(\sin(2a)+\sin(2b)+\sin(2c)+\sin(2d))\\ | |
− | + | &=\frac{1}{2}(\sin(\varphi_A+\varphi_B-\varphi_C)+\sin(\varphi_B+\varphi_C-\varphi_A)+\sin(\varphi_C+\varphi_A-\varphi_B)-\sin(\varphi_A+\varphi_B+\varphi_C))\\ | |
− | + | &=\frac{1}{2}(2\sin\varphi_B\cos(\varphi_A-\varphi_C)-2\sin\varphi_B\cos(\varphi_A+\varphi_C))\\ | |
− | + | &=\frac{1}{2}\cdot2\cdot2\sin\varphi_A\sin\varphi_B\sin\varphi_C\\ | |
+ | &=2\sin\varphi_A\sin\varphi_B\sin\varphi_C\\ | ||
+ | &=\frac{24}{35}\\ | ||
+ | \end{align*} | ||
+ | </cmath> | ||
− | Therefore, our answer is < | + | Therefore, our answer is <math>24+35=\boxed{059}</math>. |
~Solution by eric-z | ~Solution by eric-z |
Latest revision as of 23:33, 6 January 2024
Contents
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.
Note
The solution uses
We can see that this follows because where
and
are the central angles of opposite sides.
Solution 2
Suppose the four side lengths of the quadrilateral cut out arc lengths of ,
,
, and
.
.
Therefore, without losing generality,
,
, and
yields
Because
Therefore,
Using the sum-to-product identities, our area of the quadrilateral then would be
Therefore, our answer is .
~Solution by eric-z
Solution 3
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 4 (No words)
vladimir.shelomovskii@gmail.com, vvsss
Solution 5
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
.
Video Solution by MOP 2024
~r00tsOfUnity
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.