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2009 AMC 10A Problem 12

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2009 AMC 10A Problem 12AMC 10 Step by Step

Name: Date: September 13, 2026

Triangles: Area & Pythagorean · difficulty 2 of 5 · about 2 min · position P11-15

In quadrilateral $ABCD$ , $AB = 5$ , $BC = 17$ , $CD = 5$ , $DA = 9$ , and $BD$ is an integer. What is $BD$ ?

A)
$11$
B)
$12$
C)
$13$
D)
$14$
E)
$15$

Answer

Key insight

Triangle BCD forces BD > 17 - 5 = 12 and triangle ABD forces BD < 5 + 9 = 14, so the integer BD is 13.

Solution

The diagonal $BD$ is a side of two triangles, and each gives a triangle inequality.

In $\triangle BCD$ with sides $17$ , $5$ and $BD$ : the longest side must be less than the sum of the other two, so $17 < 5 + BD$ , giving $BD > 12$ .

In $\triangle ABD$ with sides $5$ , $9$ and $BD$ : $BD < 5 + 9 = 14$ .

So $12 < BD < 14$ , and the only integer in that range is $13$ . (Both triangles then exist: $5 + 13 > 17$ and $5 + 9 > 13$ .)

The answer is $\boxed{\textbf{(C)}\ 13}$ .

Why this works

A shared side of two triangles inherits bounds from both. The lower bound comes from the triangle where $BD$ is not the longest side ( $BD$ must exceed the difference of the other two), and the upper bound from the triangle where it could be the longest. Pairing a "difference" bound with a "sum" bound is the standard way to pin down an integer length.

The trap

Using only upper bounds (BD < 22, BD < 14) and guessing among 11, 12, 13, missing the lower bound BD > 12 from triangle BCD.

Common mistakes

Using only upper bounds (BD < 22, BD < 14) and guessing among 11, 12, 13, missing the lower bound BD > 12 from triangle BCD.
Applying the triangle inequality to the whole quadrilateral (e.g. $BD < 5 + 17$ only), which is far too weak.

Techniques

Bound the quantity above/below or estimate to pin it down