AMC 10 · 2005 · #25

Grade 6 geometry-2d

area-trianglesratio-proportion identify-subproblemscomplementary-counting ↑ Prerequisites: area-trianglesratio-proportionfraction-multiplication

📏 Medium solution 💡 2 insights

Problem

In triangle ABC the sides are AB=25, BC=39, AC=42. Point D sits on side AB with AD=19, and point E sits on side AC with AE=14. Segment DE cuts the triangle into a small corner triangle ADE and the four-sided region BCED that is left. Find the ratio of the area of triangle ADE to the area of quadrilateral BCED.

Pick an answer.

(A)

$\frac{266}{1521}$

(B)

$\frac{19}{75}$

(C)

$\frac{1}{3}$

(D)

$\frac{19}{56}$

(E)

AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The quadrilateral BCED has no tidy area formula, but it is just the whole triangle with the corner triangle sliced off, so the only number worth chasing is [ADE]/[ABC]; the quadrilateral is its complement (Tool #16). Comparing [ADE] to [ABC] directly is awkward because both a base and a height change at once. Tool #7 (Identify Subproblems) splits that comparison into two easy links through a middle triangle ABE: sliding B back to E changes only one dimension at a time. Each link uses the single fact that triangles with the same height have areas in the ratio of their bases, which a diagram (Tool #1) makes visible. Multiplying the two links gives [ADE]/[ABC], and the complement finishes the ratio.

1STEP 1

See the quadrilateral as a leftover

Segment DE slices the corner triangle ADE off, so the quadrilateral BCED is exactly the whole triangle minus that corner.

[BCED]=[ABC]-[ADE]

2STEP 2

Route through a middle triangle

Going from ADE straight to ABC moves base and far vertex at once, so route through middle triangle ABE, which shares a base line with each.

[ADE]/[ABC]=[ADE]/[ABE]·[ABE]/[ABC]

3STEP 3

First link: same height from E

ADE and ABE share far vertex E with bases on line AB, so equal heights make areas follow bases: $\frac{19}{25}$ .

[ADE]/[ABE]=AD/AB=19/25

4STEP 4

Second link: same height from B

ABE and ABC share far vertex B with bases on line AC, so their areas follow AE to AC: $\frac{14}{42}$ , that is $\frac{1}{3}$ .

[ABE]/[ABC]=AE/AC=14/42=1/3

5STEP 5

Multiply the two links

Multiply the two links: the corner triangle is $\frac{19}{75}$ of the whole. BC=39 never appeared — only how far D and E sit along each side.

[ADE]/[ABC]=19/25·1/3=19/75

6STEP 6

Take the complement and divide

The quadrilateral is the rest, $1-\frac{19}{75}=\frac{56}{75}$ ; dividing the two shares cancels the 75 and leaves $\frac{19}{56}$ , choice (D).

[BCED]/[ABC]=1-19/75=56/75 → [ADE]/[BCED]=19/75/56/75=19/56 → (D)

Answer

19/56

The corner triangle is 19/75≈ 0.25 of the whole triangle, so the quadrilateral holds the other ≈ 0.75 — the leftover really should be the bigger piece, and the answer 19/56≈ 0.34 correctly shows the triangle is well under half the quadrilateral. It is also a useful check to reject the near-miss choices: (B) 19/75 is [ADE]/[ABC], the ratio to the whole triangle rather than to the quadrilateral, and (A) 266/1521=(19·14)/(39·39) comes from wrongly dividing by BC; only (D) compares the two regions the problem actually names.

💡Key takeaway

Triangles with the same height compare by their bases, so slide through the middle triangle ABE to get [ADE]/[ABC]=19/25·1/3=19/75, then the leftover quadrilateral is 56/75, giving 19/56.

See the quadrilateral as a leftover
Route through a middle triangle
First link: same height from E
Second link: same height from B
Multiply the two links
Take the complement and divide