AMC 8 · 2017 · #16

Grade 6 geometry-2dalgebra

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

perimeterarea-trianglesratio-proportionpythagorean-theorem identify-subproblemsconvert-to-algebraratio-proportion ↑ Prerequisites: area-trianglespythagorean-theorem

📏 Medium solution 💡 4 insights 📊 Diagram

Problem

A right triangle ABC has legs AC = 3 and AB = 4 and hypotenuse BC = 5. Place a point D somewhere on segment BC so that the two smaller triangles formed, △ ACD and △ ABD, have the same perimeter. Find the area of △ ABD.

Pick an answer.

(A)

$frac{3}{4}$

(B)

$frac{3}{2}$

(C)

(D)

$frac{12}{5}$

(E)

$frac{5}{2}$

AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The big question "area of △ ABD" breaks cleanly into three smaller questions (Tool #7): (a) where is D on BC? (b) what is the area of the whole right triangle △ ABC? (c) what fraction of that area belongs to △ ABD? Tool #1 (Draw a Diagram) supports (a): on the given figure we label BD and CD and notice the shared side AD cancels from both perimeters, so the perimeter condition becomes a simple statement about BD and CD alone. Tool #6 (Guess and Check) handles "two pieces sum to 5, differ by 1" without needing formal algebra — perfect for an elementary solver.

1STEP 1

Draw the figure, put D on BC, and label the pieces BD and CD — notice both small triangles share the cevian AD.

BD + CD = 5

2STEP 2

Write both perimeters; the shared AD cancels, leaving AC + CD = AB + BD, i.e. 3 + CD = 4 + BD, so CD = BD + 1.

AC + CD + AD = AB + BD + AD → 3 + CD = 4 + BD → CD - BD = 1

3STEP 3

From BD + CD = 5 and CD = BD + 1, guess and check: BD = 2 and CD = 3 gives 2 + 3 = 5 with difference 1.

BD = 2, CD = 3

4STEP 4

The whole right triangle has its two legs as base and height, so its area is $\frac{1}{2}$ · 4 · 3 = 6.

Area(△ ABC) =

\frac{1}{2}

· AB · AC =

\frac{1}{2}

· 4 · 3 = 6

5STEP 5

△ ABD and △ ABC share the altitude from A, so their areas are in the ratio of the bases: BD : BC = 2 : 5.

\frac{Area(△ ABD)}{Area(△ ABC)}

\frac{BD}{BC}

\frac{2}{5}

6STEP 6

Multiply: Area(△ ABD) = $\frac{2}{5}$ · 6 = $\frac{12}{5}$ , which is choice (D).

Area(△ ABD) =

\frac{2}{5}

· 6 =

\frac{12}{5}

→ (D)

Answer

\frac{12}{5}

Sanity check the size of the answer. The whole right triangle has area 6. Since BD = 2 is less than half of BC = 5, triangle △ ABD should have less than half the total area, i.e. less than 3. We got

\frac{12}{5}

= 2.4, which is indeed less than 3 but still a meaningful chunk — exactly

\frac{2}{5}

of 6. Also

\frac{12}{5}

+ Area(△ ACD) = 6, so Area(△ ACD) =

\frac{18}{5}

, and the ratio

\frac{12}{5}

\frac{18}{5}

= 2 : 3 = BD : CD, confirming the shared-altitude argument.

💡Key takeaway

This AMC 8 problem only needs Grade 6 ratio reasoning — when two triangles share a height, their areas split in the same ratio as their bases — that you already know!

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