AMC 10 · 2007 · #22

Grade 10 geometry-2d

midpoint-formulasimilar-figuresdilation coordinate-geometryidentify-subproblems ↑ Prerequisites: midpoint-formulaarea-triangles

📏 Long solution 💡 4 insights

Problem

Two particles run the same closed triangular route at equal speed from different starts. The midpoint of the segment joining them traces a closed path. Find the ratio of the enclosed area to the triangle's.

Pick an answer.

(A)

$\frac{1}{16}$

(B)

$\frac{1}{12}$

(C)

$\frac{1}{9}$

(D)

$\frac{1}{6}$

(E)

$\frac{1}{4}$

How to solve

Strategy Introduce a Variable

The hard part is not arithmetic, it is that the problem hands us a mystery curve. So refuse to guess its shape. Introduce a clock t and one function f(t) describing the route (Tool #4), which turns both particles into formulas and the tracked point into M(t). Two structural facts then fall out for free. First, M repeats after half a lap (Tool #5), which bounds how much path there is. Second, M can only bend when a particle turns a corner, so split the time interval at those corners (Tool #7): the count of corner times is the count of straight pieces, and that is what proves the region is a triangle instead of something curved. Only then put in coordinates (Tool #1) and measure. Finally, redo the whole computation in a deliberately easier triangle (Tool #9), which both checks the number and shows which hypotheses actually mattered.

1STEP 1

One clock, one route function

One route function and a time shift describe both particles.

P(t)=f(t), Q(t)=f (t+3/2), M(t)=(P(t)+Q(t))/2

2STEP 2

The trace closes after half a lap

The traced path closes after only half a lap.

M (t+3/2)=(f (t+3/2)+f(t+3))/2=(f (t+3/2)+f(t))/2=M(t)

3STEP 3

Between corners the midpoint goes straight

Between corners the midpoint runs straight, so the path is a triangle.

corner times in [0,3/2]: t=0, 1/2, 1 ⟹ 3 straight arcs, closing at t=3/2

4STEP 4

Locate the three corners

Locating its three corners pins the shape down.

M(0)=(3/8,√(3)/8), M (1/2)=(1/2,√(3)/4), M(1)=(5/8,√(3)/8)

5STEP 5

It is a quarter-scale copy

It is a quarter-scale copy, so the ratio is 1/16.

side(R)=1/4, [R]/([△ ABC])=(1/4)²=1/16

6STEP 6

Recheck in a stretched triangle

A stretched triangle gives the same ratio, so the answer is 1/16, choice (D).

[R]/([△ ABC])=(1/2·1/4·1/4)/1/2=1/32/1/2=1/16

Answer

1/16

Every point of the path is the midpoint of two points of △ ABC, so the path lies inside the triangle and the ratio must be below 1; 1/16 passes. A sharper structural check uses rotation. Advancing the clock by exactly 1 sends A→ B→ C→ A, so f(t+1)=ρ(f(t)) where ρ is the 120° rotation about the center O of △ ABC; averaging gives M(t+1)=ρ(M(t)), so R must be carried onto itself by ρ. A triangle fixed by a 120° rotation has to be equilateral and centered at O, and both hold: the three corners average to (1/2,√(3)/6), which is exactly the centroid of △ ABC, and all three sides came out 1/4. The distractors are the natural slips. Choice (E) 1/4 is the medial triangle, what you get from believing the path passes through the three side midpoints; it does not, since every point of the path sits at height at least √(3)/8 above AB while the midpoint of AB sits at height 0. Choice (C) 1/9 and choice (D) 1/6 come from guessing scale factors 1/3 or from stopping at a linear ratio instead of squaring it.

💡Key takeaway

The midpoint of two steady walkers is itself a steady walker, so it can only turn when one of them turns a corner — three turns in all, tracing a triangle exactly a quarter as wide as ABC and therefore 1/16 of its area.

One clock, one route function
The trace closes after half a lap
Between corners the midpoint goes straight
Locate the three corners
It is a quarter-scale copy
Recheck in a stretched triangle