2021 AMC 10A Spring Problem 17

Question

Trapezoid

ABCD

has

\overline{AB}\parallel\overline{CD},

BC=CD=43,

and

\overline{AD}\perp\overline{BD}.

Let

O

be the intersection of the diagonals

\overline{AC}

and

\overline{BD},

and let

P

be the midpoint of

\overline{BD}.

Given that

OP=11,

the length of

AD

can be written in the form

m\sqrt{n},

where

m

and

n

are positive integers and

n

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

m+n?

65

132

157

194

215

Answer: D

Concepts:trapezoid similarity Pythagorean Theorem

Difficulty rating: 1950

Small hint:

Use the isosceles triangle to get a right triangle involving the midpoint of

BD

Big hint:

The intersection point

O

splits the diagonals in the ratio of the parallel bases

Video solution:

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Written solution:

Because

BC=CD,

the median from

C

to

BD

is perpendicular to

BD.

Thus

\triangle BPC

is a right triangle. Let

\angle DBC=\alpha.

Since

AB\parallel CD,

we also have

\angle ABD=\alpha,

so

\triangle BPC\sim\triangle BDA.

Since

P

is the midpoint of

BD,

we have

\frac{BD}{BP}=2.

In the similarity,

BC

corresponds to

AB,

so

\begin{aligned} \frac{AB}{BC} &=2, \\ AB &=2\cdot43=86. \end{aligned}

Also,

\triangle ABO\sim\triangle CDO,

so

\frac{BO}{OD}=\frac{AB}{CD}=2.

Since

OP=11

and

P

is the midpoint of

BD,

write

BP=PD=t.

Then

BO=t+11

and

OD=t-11,

so

\frac{t+11}{t-11}=2.

This gives

t=33,

hence

BD=66.

Finally,

\triangle ABD

is right, so

\begin{aligned} AD &= \sqrt{AB^2-BD^2} \\ &= \sqrt{86^2-66^2} \\ &= 4\sqrt{190}. \end{aligned}

Thus

m+n=4+190=194.

Thus, D is the correct answer.

2021 AMC 10A Spring Problem 17

Answer: D

Difficulty rating: 1950

Small hint:

Big hint:

Video solution:

Written solution:

Problem 17 in other years