2021 AIME II Problem 12

Question

A convex quadrilateral has area

30

and side lengths

5,

6,

9,

and

7,

in that order. Denote by

\theta

the measure of the acute angle formed by the diagonals of the quadrilateral. Then

\tan \theta

can be written in the form

\frac{m}{n},

where

m

and

n

are relatively prime positive integers. Find

m + n.

Answer

Answer: 47

Concepts:law of cosines area trigonometry

Difficulty rating: 2920

Small hint:

If the diagonals have lengths

p

and

q

and meet at angle

\theta,

the area of the quadrilateral is

\frac{1}{2}pq\sin\theta

Big hint:

The law of cosines in the four corner triangles gives

BC^2 + DA^2 - AB^2 - CD^2

= 2pq\cos(\angle APB),

where

P

is the diagonals’ intersection

Solution:

Label the quadrilateral

ABCD

with

AB = 5,

BC = 6,

CD = 9,

DA = 7,

and let the diagonals meet at

P,

cutting

\overline{AC}

into

p_1, p_2

and

\overline{BD}

into

q_1, q_2.

With

\varphi = \angle APB,

the law of cosines in the four corner triangles (whose angles at

P

alternate between

\varphi

and

180^\circ - \varphi

) gives

\begin{aligned} &BC^2 + DA^2 - AB^2 - CD^2 \\ &= 2(p_1q_1 + p_1q_2 + p_2q_1 + p_2q_2) \\ &\quad {}\cdot \cos\varphi \\ &= 2\,AC \cdot BD \cos\varphi. \end{aligned}

The left side is

36 + 49 - 25 - 81 = -21,

so

AC \cdot BD\,|\cos\varphi| = \frac{21}{2},

and the acute angle

\theta

between the diagonals satisfies

AC \cdot BD \cos\theta = \frac{21}{2}.

Meanwhile the four corner triangles give the area

\frac{1}{2} AC \cdot BD \sin\theta = 30,

so

AC \cdot BD \sin\theta = 60.

Dividing,

\tan\theta = \frac{60}{\frac{21}{2}} = \frac{40}{7},

so

m + n = 40 + 7 = 47.