Difference between revisions of "2021 AIME II Problems/Problem 12"
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==Problem== | ==Problem== | ||
A convex quadrilateral has area <math>30</math> and side lengths <math>5, 6, 9,</math> and <math>7,</math> in that order. Denote by <math>\theta</math> the measure of the acute angle formed by the diagonals of the quadrilateral. Then <math>\tan \theta</math> can be written in the form <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. | A convex quadrilateral has area <math>30</math> and side lengths <math>5, 6, 9,</math> and <math>7,</math> in that order. Denote by <math>\theta</math> the measure of the acute angle formed by the diagonals of the quadrilateral. Then <math>\tan \theta</math> can be written in the form <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. | ||
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==Solution 1 (Sines and Cosines)== | ==Solution 1 (Sines and Cosines)== |
Revision as of 00:01, 7 September 2021
Contents
Problem
A convex quadrilateral has area and side lengths
and
in that order. Denote by
the measure of the acute angle formed by the diagonals of the quadrilateral. Then
can be written in the form
, where
and
are relatively prime positive integers. Find
.
Solution 1 (Sines and Cosines)
Solution 1.1 (Version 1)
We denote by ,
,
and
four vertices of this quadrilateral, such that
,
,
,
.
We denote by
the point that two diagonals
and
meet at.
To simplify the notation, we denote
,
,
,
.
We denote . Hence,
and
.
First, we use the triangle area formula with sines to write down an equation of the area of the quadrilateral .
We have
where the second equality follows from the formula to use the sine function to compute a triangle area, the the fourth equality follows from the property that
.
Because , we have
Second, we use the law of cosines to establish four equations for four sides of the quadrilateral
.
In , following from the law of cosines, we have
Because
and
, we have
In
, following from the law of cosines, we have
Because
and
, we have
In
, following from the law of cosines, we have
Because
and
, we have
In
, following from the law of cosines, we have
Because
and
, we have
By taking
, we get
By taking
, we get
Therefore, by writing this answer in the form of
, we have
and
.
Therefore, the answer to this question is .
~ Steven Chen (www.professorchenedu.com)
Solution 1.2 (Version 2)
Since we are asked to find , we can find
and
separately and use their values to get
. We can start by drawing a diagram. Let the vertices of the quadrilateral be
,
,
, and
. Let
,
,
, and
. Let
,
,
, and
. We know that
is the acute angle formed between the intersection of the diagonals
and
.
We are given that the area of quadrilateral
is
. We can express this area using the areas of triangles
,
,
, and
. Since we want to find
and
, we can represent these areas using
as follows:
We know that
. Therefore it follows that:
From here we see that
. Now we need to find
. Using the Law of Cosines on each of the four smaller triangles, we get following equations:
We know that
. We can substitute this value into our equations to get:
If we subtract the sum of the first and third equation from the sum of the second and fourth equation, the squared terms cancel, leaving us with:
From here we see that
.
Since we have figured out and
, we can calculate
:
Therefore our answer is
.
~ my_aops_lessons
Solution 2 (Right Triangles)
This solution refers to the Diagram section.
In convex quadrilateral let
and
Let
and
be the feet of the perpendiculars from
and
respectively, to
We obtain the following diagram:
Let and
We apply the Pythagorean Theorem to right triangles
and
respectively:
Let the brackets denote areas. We get
We subtract
from
From right triangles
and
we have
It follows that
Finally, we divide
by
from which the answer is
~MRENTHUSIASM
Solution 3
Let ,
,
,
be the vertices of the quadrilateral,
,
,
,
be the lengths of the sides of
, and
and
be the lengths of the diagonals of
. By Bretschneider's formula,
. Thus,
. Also,
; solving for
yields
. Since
is acute,
is positive, so
. Solving for
yields
, for a final answer of
.
~ Leo.Euler
Video Solution
https://www.youtube.com/watch?v=7DxIdTLNbo0
See Also
2021 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.