AMC 10 · 2002 · #23

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity
median-of-trianglepythagorean-theoremmidpoint-formula coordinate-geometryconvert-to-algebra ↑ Prerequisites: pythagorean-theoremmedian-of-triangle
📏 Long solution 💡 3 insights
Problem
Triangle ABC has two known sides: AB = 1 and AC = 2. Mark the midpoint of the third side BC and draw the segment from A to that midpoint — this is the median from A. The condition is that this median is exactly as long as BC itself. Find the length BC.

Pick an answer.

(A)
$\frac{1+\sqrt{2}}{2}$
(B)
$\frac{1+\sqrt{3}}2$
(C)
$\sqrt{2}$
(D)
$\frac 32$
(E)
$\sqrt{3}$
How to solve
Strategy Introduce a Variable

Only one number is really unknown, so tool #4 (Introduce a Variable) starts it: write BC = 2m so the two halves are m each and the median is 2m. Then tool #1 (Draw a Diagram) is upgraded into a coordinate picture — park the midpoint M at the origin with BC along the x-axis. That placement is the whole trick, because it makes B and C sit at (-m, 0) and (m, 0), perfectly symmetric about the origin. Tool #7 (Identify Subproblems) finishes it: write AB², AC² and AM² separately, then notice that adding the first two wipes out the position of A along the x-axis and leaves exactly the quantity AM². Coordinates are chosen over an altitude-and-Pythagoras picture on purpose: an altitude may land inside or outside BC, and guessing which would be an assumption we have not earned.

1STEP 1

Name the half-side

Write BC = 2m so the midpoint halves give BM = m and the condition reads AM = 2m.

BC = 2m, BM = MC = m, AM = 2m
2STEP 2

Park the midpoint at the origin

Put the midpoint at the origin with BC on an axis; rigid motion changes no length.

M = (0,0), B = (-m, 0), C = (m, 0), A = (p, q)
3STEP 3

Write the three lengths

The distance formula gives three equations in p, q and m.

(p+m)² + q² = 1, (p-m)² + q² = 4, p² + q² = 4m²
4STEP 4

Add the first two equations

Adding the two side equations cancels the cross terms, leaving 2(p² + q²) + 2m² = 5.

2(p² + q²) + 2m² = 5
5STEP 5

Substitute the median and solve

That bundle is the median squared, so 10m² = 5 and BC = √(2), choice (C).

10m² = 5 → m = √(2)/2 → BC = 2m = √(2)
Answer
√(2)
Build the triangle and measure it. Subtracting the first two equations gives (p+m)² - (p-m)² = 4pm = 1 - 4 = -3, so with m = √(2)/2 we get p = -3/4m = -3√(2)/4 ≈ -1.0607, and then q² = 4m² - p² = 2 - 9/8 = 7/8, so q ≈ 0.9354. Since q² ＞ 0, the triangle really exists. Checking: AB² = (-1.0607 + 0.7071)² + 0.875 = 0.125 + 0.875 = 1 and AC² = (-1.0607 - 0.7071)² + 0.875 = 3.125 + 0.875 = 4 and AM² = 1.125 + 0.875 = 2 = BC² — all three conditions hold. A second, cheaper sanity check: doubling the median past M builds a triangle with sides AB = 1, AC = 2 and third side 2 · AM, so the triangle inequality forces 2 · AM to sit strictly between 2 - 1 = 1 and 2 + 1 = 3, that is 1/2 ＜ BC ＜ 3/2. That range excludes 3/2 and √(3) ≈ 1.732 outright, and √(2) ≈ 1.414 fits comfortably inside it.
💡Key takeaway

Put the midpoint at the origin: the two ends of the side land at -m and +m, so adding the two distance equations cancels everything about where the top vertex leans and leaves the median staring back at you.

  • Name the half-side
  • Park the midpoint at the origin
  • Write the three lengths
  • Add the first two equations
  • Substitute the median and solve

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