AMC 10 · 2004 · #18

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

area-trianglesratio-proportion complementary-countingidentify-subproblems ↑ Prerequisites: area-triangles

📏 Long solution 💡 4 insights 📊 Diagram

Problem

In right triangle ACE with AC=12, CE=16, EA=20, a point sits on each side: B on AC with AB=3, D on CE with CD=4, and F on EA with EF=5. Find the ratio of the area of the inner triangle DBF to the area of ACE.

Pick an answer.

(A)

$\frac{1}{4}$

(B)

$\frac{9}{25}$

(C)

$\frac{3}{8}$

(D)

$\frac{11}{25}$

(E)

$\frac{7}{16}$

AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Change Focus / Count the Complement

Chasing the inner triangle DBF head-on is awkward — none of its sides lie along the big triangle. But the three cut-off corners are easy: each corner triangle shares a full angle with ACE. So change focus (Tool #16): find the three corner areas and subtract them from the whole. The corner at C is a right triangle you can measure directly; the corners at A and E come from a same-height ratio argument. Breaking the figure into whole = inner + three corners is the subproblem split (Tool #7), read straight off the drawing (Tool #1).

1STEP 1

Locate the right angle at C

Since 12²+16²=20², the right angle sits at C, so the legs AC and CE give area 96.

12²+16²=20² → ∠ C = 90°, [△ ACE]=1/2 · 12 · 16 = 96

2STEP 2

Whole equals inner plus three corners

Each point cuts a corner triangle off ACE, so DBF is exactly what is left after removing ABF, BCD and DEF.

[△ DBF] = [△ ACE] - [△ ABF] - [△ BCD] - [△ DEF]

3STEP 3

Corner at C is a right triangle

BCD keeps the right angle at C, with legs BC=9 and CD=4, so its area is 18 — that is 3/16 of 96.

[△ BCD]=1/2 · 9 · 4 = 18, ([△ BCD])/([△ ACE])=18/96=3/16

4STEP 4

Corner at A by a same-height ratio

ABF shares angle A with ACE, so shrinking AC to AB scales by 3/12 and AE to AF by 15/20 — the product is 3/16.

([△ ABF])/([△ ACE])=AB/AC·AF/AE=1/4·3/4=3/16

5STEP 5

Corner at E the same way

DEF shares angle E, and the same argument gives 12/16 times 5/20 — again 3/16, so all three corners match.

([△ DEF])/([△ ACE])=ED/EC·EF/EA=3/4·1/4=3/16

6STEP 6

Subtract the three corners

Three corners take 3/16 each, so together 9/16; the inner triangle DBF gets 1-9/16 = 7/16, choice (E).

([△ DBF])/([△ ACE]) = 1-9/16=7/16 → (E)

Answer

7/16

The inner triangle clearly sits well inside ACE, so its share should be under 1/2, and 7/16 (just below one half) fits the picture. In real numbers the three corners are 18+18+18=54, and 96-54=42, giving 42/96=7/16 — the same result. The trap choice 9/25 would come from wrongly using (3/5)²-type reasoning, and 1/4 from forgetting the corners are unequal-looking, but the honest corner areas all equal 3/16 and give 7/16.

💡Key takeaway

When a triangle is trimmed at each corner, find the easy corner pieces and subtract them from the whole — two triangles that share an angle have areas in the ratio of the two side lengths multiplied together.

Locate the right angle at C
Whole equals inner plus three corners
Corner at C is a right triangle
Corner at A by a same-height ratio
Corner at E the same way
Subtract the three corners