Competition · AMC preparation · step 4 of 4

AMC 10 · 2019B · #16

Grade 8 geometry-2d

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

pythagorean-theoremisosceles-triangleinteger-pythagorean-triplesangle-sum-trianglesimilar-triangles identify-subproblemscasework ↑ Prerequisites: pythagorean-theoremisosceles-triangleangle-sum-triangle

📏 Long solution 💡 4 insights

Problem

In right triangle ABC with the right angle at C, point D is on segment AB and point E is on segment BC. We are told AC = CD, DE = EB, and AC : DE = 4 : 3. Find the ratio AD : DB.

Pick an answer.

(A)

2:3

(B)

$2:\sqrt{5}$

(C)

1:1

(D)

$3:\sqrt{5}$

(E)

3:2

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 (Diagram): the problem mixes angles, equal sides, and a point on AB — a labeled picture is essential. Tool #7 (Subproblems): split into (a) find BC using the two isosceles triangles, (b) drop altitudes from C and E onto AB to read off AD and DB. Tool #9 (Easier Problem): pick concrete numbers AC = 4 and DE = 3 so we work with integers instead of a ratio.

1STEP 1

Pick concrete side lengths

Replace the ratio with concrete lengths: let AC = 4 and DE = 3, so CD = 4 and EB = 3.

AC = CD = 4, DE = EB = 3

2STEP 2

Find the right angle at D

Both isosceles triangles share their base angles with △ ABC, so ∠ ADC + ∠ EDB = 90°; the leftover angle at D is ∠ CDE = 90°.

∠ CDE = 90°

3STEP 3

Find CE with the Pythagorean theorem

So △ CDE is right-angled at D with legs 4 and 3 — the hypotenuse is CE = 5 (a 3–4–5 triangle).

CE = √(4² + 3²) = 5

4STEP 4

Compute BC and AB

Since E lies between B and C, BC = 8 from CE + EB, and AB = √(16 + 64) = 4√(5).

BC = 8, AB = 4√(5)

5STEP 5

Find AD

Drop the altitude from C in isosceles △ ACD; it bisects AD, giving AD = 2·AC·cos∠CAB = 2·4·( $\frac{1}{\sqrt{5}}$ ) = $\frac{8}{\sqrt{5}}$ .

AD = 2 · AC · AC/AB = (2 · 16)/4√(5) = 8/√(5)

6STEP 6

Find DB the same way

Mirror it in isosceles △ DEB: the altitude from E bisects DB, so DB = 2·EB·cos∠EBD = 2·3·( $\frac{2}{\sqrt{5}}$ ) = $\frac{12}{\sqrt{5}}$ .

DB = (2 · 3 · 2)/√(5) = 12/√(5)

7STEP 7

Form the ratio AD to DB

The √(5) denominators cancel: AD : DB = 8 : 12 = 2 : 3, which is choice (A).

AD : DB = 8 : 12 = 2 : 3

Answer

2:3

Quick sanity: AD + DB should equal AB. AD + DB =

\frac{8}{\sqrt{5}}

\frac{12}{\sqrt{5}}

\frac{20}{\sqrt{5}}

= 4√(5), which matches AB = 4√(5). The ratio 2 : 3 also makes geometric sense — D is closer to A than to B, which fits the picture where AC = 4 ＜ BC = 8 so the larger leg's foot of the cevian-from-apex sits farther along AB.

💡Key takeaway

This AMC 10 problem only needs Grade 8 Pythagorean-theorem reasoning you already know — spot the hidden 3–4–5 right triangle at D, then split the big triangle into two isosceles pieces and read off AD : DB = 2 : 3. The answer is (A).

Pick concrete side lengths
Find the right angle at D
Find CE with the Pythagorean theorem
Compute BC and AB
Find AD
Find DB the same way
Form the ratio AD to DB

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