AMC 10 · 2012 · #18
Grade 10 geometry-2dPick an answer.
The segment BI is not a side, a median, or an altitude, so nothing in the given figure measures it directly. Tool #1 (Draw a Diagram) supplies the missing object: drop the perpendicular from I to each side. Because I sits on all three bisectors, those three perpendiculars have one common length r, and BI instantly becomes the hypotenuse of a right triangle with a known leg r. That converts the problem into Tool #7 (Identify Subproblems): find the other leg, find r, then combine with the Pythagorean Theorem. Tool #4 (Introduce a Variable) handles the first subproblem — naming the three "touch distances" x,y,z turns the three side lengths into a tiny linear system. Tool #15 (Organize Information in More Ways) handles the second — writing the area of the triangle two different ways, once from the three sides and once as three triangles sharing the apex I, pins r down. The care point throughout is that the picture must be proved, not assumed: the concurrency of the bisectors, the equality of the two touch distances at each vertex, and the fact that the perpendicular feet land inside the sides are the three claims the whole computation rests on.
Prove what the point I really is
The point is the one equally far from all three sides.
An angle bisector is exactly the set of points equally far from the angle's two sides, so crossing two bisectors forces equal distance to all three sides at once.
An angle bisector is exactly the set of points equally far from the angle's two sides.
▸ Why?
Those equidistant points make up the line that folds one side of the angle onto the other.
▸ Why?
Distance to a side is measured square on, so those equal distances are radii of one touching circle.
Drop perpendiculars and match them in pairs
Tangent lengths from each corner come in equal pairs.
Two right triangles that share a hypotenuse and have the same leg are copies of each other, so the two touch distances at a vertex must agree.
10.G-SRT.B.5Draw A DiagramSolve for the touch distance at B
That gives the corner's touch distance, 13.
Adding all three side equations gives the grand total once; then each unknown pops out by subtracting the side that happens to miss it.
8.EE.C.8Introduce A VariableGet the area from the three sides
The three sides give the area directly.
Three side lengths pin a triangle down completely, so its area is already decided before any other choice is made.
7.G.B.6Identify SubproblemsCount the same area a second way
Counting the area again gives the inner radius.
Writing one area in two different ways turns the unknown height r into an equation with a single solution.
6.G.A.1Organize Information In More WaysAssemble BI from its two legs
Assembling the right triangle gives 15, choice (A).
Once the incenter is described by how far it sits along a side and how far it sits off that side, the distance to the vertex is just a right-triangle hypotenuse.
8.G.B.7Identify SubproblemsTo reach the incenter from a corner, go along a side by the touch distance and then straight off the side by the inradius — those two are the legs of a right triangle, so here BI = √(13² + 56) = 15.
- Prove what the point I really is
- Drop perpendiculars and match them in pairs
- Solve for the touch distance at B
- Get the area from the three sides
- Count the same area a second way
- Assemble BI from its two legs