AMC 10 · 2009 · #12
Grade 7 geometry-2d
Pick an answer.
AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The diagonal BD belongs to two different triangles at once, so Tool #7 (Identify Subproblems) splits the four-sided figure into △ BCD (sides 17, 5, BD) and △ ABD (sides 5, 9, BD). Inside each triangle the third side cannot be so long or so short that the other two fail to close up — that boundary is exactly the triangle inequality, so Tool #14 (Extreme Principle) pushes each triangle to its flat, zero-area edge to read off how big and how small BD is allowed to be. Those two boundaries trap BD inside a narrow band, and Tool #3 (Eliminate Possibilities) uses the whole-number clue to pick the single integer that survives.
Cut the quadrilateral into two triangles
Draw diagonal BD: it makes △ BCD with sides 17, 5, BD and △ ABD with sides 5, 9, BD, which must fit both triangles at once.
One diagonal turns a hard four-sided figure into two triangles that share that diagonal as a common side.
7.G.A.2Identify SubproblemsLower bound from the big triangle
In △ BCD, sides 5 and BD must out-reach 17, so BD is greater than 12; at exactly 12 they lie flat on the 17.
Two sides can only wrap around a long third side if together they are longer than it — otherwise they collapse into a straight line.
Two sides can only wrap around a long third side if together they outreach it.
▸ Why?
Otherwise the ends never meet and the triangle collapses into a straight line.
▸ Why?
Those comparisons chain together, trapping the unknown length between a floor and a ceiling.
Upper bound from the small triangle
In △ ABD, BD cannot beat 5+9, so BD is less than 14; at exactly 14 the two sides lie flat end to end.
A side can be at most as long as the other two laid end to end — reach that length and the triangle flattens to a segment.
6.EE.B.8Extreme PrincipleTrap the integer
Together they trap BD strictly between 12 and 14, and the only integer there is 13 — choice (C).
Squeeze a value between two consecutive-ish bounds and the whole-number clue leaves exactly one survivor.
6.NS.C.7Eliminate PossibilitiesA diagonal cuts a quadrilateral into two triangles; make each triangle just barely close to trap the diagonal between two numbers, then the whole-number clue picks the winner.
- Cut the quadrilateral into two triangles
- Lower bound from the big triangle
- Upper bound from the small triangle
- Trap the integer