AMC 10 · 2009 · #14

Grade 10 geometry-2d
coordinate-geometrymedian-of-trianglequadratic-equations convert-to-algebraidentify-subproblems ↑ Prerequisites: coordinate-geometry
📏 Medium solution 💡 3 insights
Problem
A line's slope also decides where a triangle's third corner sits. The line must split the triangle into two equal areas. Find the sum of the possible slopes.

Pick an answer.

(A)
$- \!\frac {1}{3}$
(B)
$- \!\frac {1}{6}$
(C)
$\frac {1}{6}$
(D)
$\frac {1}{3}$
(E)
$\frac {1}{2}$
How to solve
Strategy Draw a Diagram

Sketching the triangle shows the one fact that cracks the problem: the line y = mx passes through the origin, and the origin is a corner of the triangle. So this is not some random slice — it is a cut from a vertex to the opposite side, and for that kind of cut "equal areas" has a clean geometric meaning. Turn that meaning into coordinates, get one quadratic in m, solve it, and then do the step the equation cannot do for you: put each value of m back into the picture and confirm it really builds a triangle that really gets halved. Only values that survive that test are allowed into the sum.

1STEP 1

The cutting line runs through a corner

The line passes through one corner no matter what.

A = (0,0) satisfies y = mx for all m; C = (6m,0) = A ⇔ m = 0
2STEP 2

Equal areas means D is the midpoint

So equal areas means it hits the opposite midpoint.

[ABD] = 1/2 · BD · h, [ACD] = 1/2 · DC · h ⟹ [ABD] = [ACD] ⇔ BD = DC
3STEP 3

Force the line through that midpoint

Forcing it through gives one quadratic.

M = ( (1+6m)/2, 1/2 ), 1/2 = m · (1+6m)/2 ⟹ 6m² + m - 1 = 0
4STEP 4

Solve the quadratic

It factors into two roots.

6m² + m - 1 = (3m-1)(2m+1) = 0 ⟹ m = 1/3 or m = -1/2
5STEP 5

Put both values back in the picture

Both give genuine triangles, so both survive.

m = 1/3: C=(2,0), M=(3/2,1/2), [ABC]=1; m = -1/2: C=(-3,0), M=(-1,1/2), [ABC]=3/2
6STEP 6

Add the surviving values

Adding gives -1/6, choice (B).

1/3 + ( -1/2 ) = 2/6 - 3/6 = -1/6
Answer
- 1/6
Check the halving directly with coordinates instead of trusting the midpoint rule. For m = 1/3 the pieces are A(0,0), B(1,1), M(3/2,1/2) and A(0,0), C(2,0), M(3/2,1/2); the shoelace formula gives 1/2 for each, and the full triangle has area 1. For m = -1/2 each piece comes out to 3/4 against a full area of 3/2. Both splits are real, so both roots count. The size of the answer is also sensible: the roots 1/3 and -1/2 straddle zero and nearly cancel, so the sum should be small and negative, and -1/6 is exactly that. Notice the trap built into the choice list: a solver who finds only the positive root stops at 1/3 and picks (D). The answer is (B) -1/6.
💡Key takeaway

A cut that starts at a corner splits a triangle into equal halves exactly when it lands on the midpoint of the opposite side, and every value the equation hands back still has to be put into the picture to prove it really builds a triangle.

  • The cutting line runs through a corner
  • Equal areas means D is the midpoint
  • Force the line through that midpoint
  • Solve the quadratic
  • Put both values back in the picture
  • Add the surviving values