2021 AMC 12B Spring Problem 24

Question

Let

ABCD

be a parallelogram with area

15.

Points

P

and

Q

are the projections of

A

and

C,

respectively, onto the line

BD;

and points

R

and

S

are the projections of

B

and

D,

respectively, onto the line

AC.

See the figure, which also shows the relative locations of these points.

Suppose

PQ=6

and

RS=8,

and let

d

denote the length of

\overline{BD},

the longer diagonal of

ABCD.

Then

d^2

can be written in the form

m+n\sqrt p,

where

m,

n,

and

p

are positive integers and

p

is not divisible by the square of any prime. What is

m+n+p?

81

89

97

105

113

Answer: A

Concepts:parallelogram trigonometry area

Difficulty rating: 2480

Small hint:

Let

\theta

be the angle between the diagonals; then

PQ=AC\cos\theta

and

RS=BD\cos\theta

Big hint:

The area is

\tfrac12\cdot AC\cdot BD\sin\theta=15,

which combines with

AC\cos\theta=6,

BD\cos\theta=8

Solution:

Let the diagonals meet at

O

at angle

\theta.

The feet of the perpendiculars from

A

and

C

to

BD

are symmetric about

O,

so

PQ=AC\cos\theta=6;

likewise

RS=BD\cos\theta=8.

The parallelogram’s area is

\tfrac12\cdot AC\cdot BD\sin\theta=15,

so

AC\cdot BD\sin\theta=30.

Then

\dfrac{48\sin\theta}{\cos^2\theta}=30,

giving

\dfrac{\sin\theta}{\cos^2\theta}=\dfrac58.

Writing

s=\sin\theta,

8s=5(1-s^2)

gives

s=\dfrac{-4+\sqrt{41}}{5},

so

\cos^2\theta=1-s^2=\dfrac{8(\sqrt{41}-4)}{25}.

Then

d^2=BD^2

=\dfrac{64}{\cos^2\theta}

=8(\sqrt{41}+4)

=32+8\sqrt{41},

so

m+n+p=32+8+41=81.

Thus, the correct answer is A.

2021 AMC 12B Spring Problem 24

Answer: A

Difficulty rating: 2480

Small hint:

Big hint:

Solution:

Problem 24 in other years