AMC 8 · 2017 · #16
Grade 6 geometry-2dalgebraProblem
In the figure below, choose point D on BC so that △ACD and △ABD have equal perimeters. What is the area of △ABD?
Pick an answer.
AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A right triangle $ABC$ has legs $AC = 3$ and $AB = 4$ and hypotenuse $BC = 5$. Place a point $D$ somewhere on segment $\overline{BC}$ so that the two smaller triangles formed, $\triangle ACD$ and $\triangle ABD$, have the same perimeter. Find the area of $\triangle ABD$.
Givens: Right triangle with legs $AC = 3$, $AB = 4$, hypotenuse $BC = 5$ (the right angle is at $A$); Point $D$ lies on $\overline{BC}$, so $BD + DC = 5$; Perimeter of $\triangle ACD$ equals perimeter of $\triangle ABD$; Answer choices: (A) $\tfrac{3}{4}$, (B) $\tfrac{3}{2}$, (C) $2$, (D) $\tfrac{12}{5}$, (E) $\tfrac{5}{2}$
Unknowns: The area of $\triangle ABD$
Understand
Restated: A right triangle $ABC$ has legs $AC = 3$ and $AB = 4$ and hypotenuse $BC = 5$. Place a point $D$ somewhere on segment $\overline{BC}$ so that the two smaller triangles formed, $\triangle ACD$ and $\triangle ABD$, have the same perimeter. Find the area of $\triangle ABD$.
Givens: Right triangle with legs $AC = 3$, $AB = 4$, hypotenuse $BC = 5$ (the right angle is at $A$); Point $D$ lies on $\overline{BC}$, so $BD + DC = 5$; Perimeter of $\triangle ACD$ equals perimeter of $\triangle ABD$; Answer choices: (A) $\tfrac{3}{4}$, (B) $\tfrac{3}{2}$, (C) $2$, (D) $\tfrac{12}{5}$, (E) $\tfrac{5}{2}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #6 Guess and Check
The big question "area of $\triangle ABD$" breaks cleanly into three smaller questions (Tool #7): (a) where is $D$ on $\overline{BC}$? (b) what is the area of the whole right triangle $\triangle ABC$? (c) what fraction of that area belongs to $\triangle ABD$? 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.
Execute — Answer: D
4.G.A.1 Step 1 - Draw the figure and add labels.
- Triangle $ABC$ has the right angle at $A$, legs $AC = 3$ and $AB = 4$, hypotenuse $BC = 5$.
- Drop point $D$ on $\overline{BC}$ and mark the two unknown pieces $BD$ and $CD$.
- Notice that both small triangles share the cevian $AD$.
💡 Labeling points, segments, and the shared cevian on the diagram is exactly the Grade 4 "identify lines and segments in figures" skill.
3.MD.D.8 Step 2 - Write each perimeter and cancel the shared side.
- The perimeter of $\triangle ACD$ is $AC + CD + AD$ and the perimeter of $\triangle ABD$ is $AB + BD + AD$.
- Setting them equal, $AD$ cancels, leaving $AC + CD = AB + BD$, i.e.
- $3 + CD = 4 + BD$, so $CD = BD + 1$.
💡 Adding up the three sides of each triangle is Grade 3 polygon-perimeter work; the cancellation is just "same thing on both sides".
4.OA.A.3 Step 3 - Find $BD$ using the two facts $BD + CD = 5$ and $CD = BD + 1$.
- Guess and check: if $BD = 2$, then $CD = 3$, and $2 + 3 = 5$ with a difference of $1$ — both conditions hold.
- So $BD = 2$ and $CD = 3$.
💡 "Two whole numbers sum to $5$ and differ by $1$" is a one-line multi-step word problem at the Grade 4 level.
6.G.A.1 Step 4 - Compute the area of the whole right triangle $\triangle ABC$ using its two legs as base and height.
- With $AB = 4$ and $AC = 3$ meeting at a right angle, the area is $\tfrac{1}{2} \cdot 4 \cdot 3 = 6$.
💡 Half of base times height for a triangle is the Grade 6 area formula.
6.RP.A.3 Step 5 - Use the shared-altitude principle.
- Triangles $\triangle ABD$ and $\triangle ABC$ both have their apex at $A$ and bases along the same line $BC$, so they share the same altitude from $A$.
- When two triangles share an altitude, the ratio of their areas equals the ratio of their bases: $\text{Area}(\triangle ABD) : \text{Area}(\triangle ABC) = BD : BC = 2 : 5$.
💡 Comparing two area-pieces that share a height is a Grade 6 ratio-reasoning move.
5.NF.B.6 Step 6 - Multiply to get the final area: $\text{Area}(\triangle ABD) = \tfrac{2}{5} \cdot 6 = \tfrac{12}{5}$.
- This matches answer choice (D).
💡 Multiplying a fraction by a whole number ($\tfrac{2}{5}$ of $6$) is a Grade 5 fraction-of-a-whole calculation.
4.G.A.1 Draw the figure and add labels. Triangle $ABC$ has the right angle at $A$, legs 3.MD.D.8 Write each perimeter and cancel the shared side. The perimeter of $\triangle ACD 4.OA.A.3 Find $BD$ using the two facts $BD + CD = 5$ and $CD = BD + 1$. Guess and check: 6.G.A.1 Compute the area of the whole right triangle $\triangle ABC$ using its two legs 6.RP.A.3 Use the shared-altitude principle. Triangles $\triangle ABD$ and $\triangle ABC$ 5.NF.B.6 Multiply to get the final area: $\text{Area}(\triangle ABD) = \tfrac{2}{5} \cdot Review
Reasonableness: 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 $\triangle ABD$ should have less than half the total area, i.e. less than $3$. We got $\tfrac{12}{5} = 2.4$, which is indeed less than $3$ but still a meaningful chunk — exactly $\tfrac{2}{5}$ of $6$. Also $\tfrac{12}{5} + \text{Area}(\triangle ACD) = 6$, so $\text{Area}(\triangle ACD) = \tfrac{18}{5}$, and the ratio $\tfrac{12}{5} : \tfrac{18}{5} = 2 : 3 = BD : CD$, confirming the shared-altitude argument.
Alternative: Tool #13 (Convert to Algebra) gives a direct route: let $BD = x$, then $CD = 5 - x$, and the perimeter condition $3 + (5 - x) = 4 + x$ collapses to $8 - x = 4 + x$, so $x = 2$. Then compute the area exactly as in steps 4-6. Algebra is cleaner here, but the guess-and-check path above is faster for an elementary student and avoids introducing a variable.
CCSS standards used (min grade 6)
4.G.A.1Draw points, lines, line segments, rays, angles, and identify in figures (Labeling $D$ on $\overline{BC}$ and marking the two new segments $BD$ and $CD$ plus the shared cevian $AD$ on the figure.)3.MD.D.8Solve real-world problems involving perimeters of polygons (Writing the perimeters of $\triangle ACD$ and $\triangle ABD$ as sums of three sides and cancelling the shared side $AD$ to get $3 + CD = 4 + BD$.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Finding two whole-number lengths that sum to $5$ and differ by $1$, giving $BD = 2$ and $CD = 3$.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Computing the area of right triangle $\triangle ABC$ as $\tfrac{1}{2} \cdot 4 \cdot 3 = 6$.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Setting the area ratio $\text{Area}(\triangle ABD) : \text{Area}(\triangle ABC)$ equal to the base ratio $BD : BC = 2 : 5$ since the two triangles share an altitude from $A$.)5.NF.B.6Solve real-world problems involving multiplication of fractions and mixed numbers (Multiplying $\tfrac{2}{5} \cdot 6 = \tfrac{12}{5}$ to extract the final area of $\triangle ABD$.)
⭐ 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!
⭐ 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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