2009 AMC 10A Problem 12

P11-15~2 minPrint

Triangles: Area & Pythagorean Bound the quantity above/below or estimate to pin it down

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

Show solution without answering

Key insight

In triangle BCD, BD + 5 must be more than 17, so BD > 12. In triangle ABD, BD < 5 + 9 = 14. So BD = 13.

Solution

The diagonal $BD$ splits the shape into two triangles. In any triangle, the two shorter sides together must be longer than the third side. Use this in both triangles.

1. Triangle $BCD$ has sides $17$ , $5$ and $BD$ . The side $17$ is long, so the other two together must beat it:
$5 + BD > 17, \qquad BD > 12.$
2. Triangle $ABD$ has sides $5$ , $9$ and $BD$ . Here $BD$ must be less than the other two together:
$BD < 5 + 9 = 14.$
3. So $BD$ is more than $12$ and less than $14$ . The only whole number that fits is $13$ .
4. Check both triangles with $BD = 13$ : $5 + 13 = 18 > 17$ and $5 + 9 = 14 > 13$ . Both are real triangles.

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

Why this works

A side shared by two triangles gets limits from both. A lower limit comes from the triangle with a long side that the shared side must help cover ( $BD$ must be more than $17 - 5$ ), and an upper limit from the other triangle ( $BD$ must be less than $5 + 9$ ). Squeezing a whole number between the two limits is a common AMC pattern.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(C)}\ 13}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Geometry, Chapter 6: Right Triangles (§6.1, §6.2, §6.3)
Introduction to Geometry, Chapter 10: Geometric Inequalities (§10.3)