AMC 10 · 2004 · #18
Grade 8 geometry-2dIn the right triangle △ACE, we have AC=12, CE=16, and EA=20. Points B, D, and F are located on AC, CE, and EA, respectively, so that AB=3, CD=4, and EF=5. What is the ratio of the area of △DBF to that of △ACE?
Pick an answer.
AMC 10 2004 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: 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 $ACE$.
Givens: Triangle $ACE$ is right-angled, with side lengths $AC=12$, $CE=16$, $EA=20$.; $B$ lies on $AC$ with $AB=3$, so $BC=12-3=9$.; $D$ lies on $CE$ with $CD=4$, so $DE=16-4=12$.; $F$ lies on $EA$ with $EF=5$, so $FA=20-5=15$.; Answer choices: (A) $\tfrac14$, (B) $\tfrac{9}{25}$, (C) $\tfrac38$, (D) $\tfrac{11}{25}$, (E) $\tfrac{7}{16}$.
Unknowns: The ratio $[\triangle DBF] : [\triangle ACE]$ of the inner triangle's area to the big triangle's area.
Understand
Restated: 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 $ACE$.
Givens: Triangle $ACE$ is right-angled, with side lengths $AC=12$, $CE=16$, $EA=20$.; $B$ lies on $AC$ with $AB=3$, so $BC=12-3=9$.; $D$ lies on $CE$ with $CD=4$, so $DE=16-4=12$.; $F$ lies on $EA$ with $EF=5$, so $FA=20-5=15$.; Answer choices: (A) $\tfrac14$, (B) $\tfrac{9}{25}$, (C) $\tfrac38$, (D) $\tfrac{11}{25}$, (E) $\tfrac{7}{16}$.
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #7 Identify Subproblems, #1 Draw a Diagram
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).
Execute — Answer: E
8.G.B.6 Step 1 Locate the right angle at C
- The triangle is right-angled, but we must find where.
- The longest side is the hypotenuse, and $EA=20$ is longest.
- Check with the side lengths: $12^2+16^2 = 144+256 = 400 = 20^2$.
- Since the two legs $AC=12$ and $CE=16$ are the ones that square-and-add to the hypotenuse, the right angle sits at $C$, between them.
- That makes $AC$ and $CE$ perpendicular, so the area of $ACE$ is half the product of those legs.
💡 The right angle always faces the longest side, so the two legs are the sides meeting at that corner.
6.G.A.1 Step 2 Whole equals inner plus three corners
- The three points cut three small triangles off the corners of $ACE$: triangle $ABF$ at vertex $A$, triangle $BCD$ at vertex $C$, and triangle $DEF$ at vertex $E$.
- Whatever is left in the middle is exactly the target triangle $DBF$.
- So instead of measuring $DBF$ directly, find the three corner areas and subtract.
- Also note the leftover lengths: $BC=9$, $DE=12$, $FA=15$.
💡 It is easier to shave off the three easy corners than to build the awkward middle piece from scratch.
6.G.A.1 Step 3 Corner at C is a right triangle
- Triangle $BCD$ shares the right angle at $C$, so it is itself a right triangle with perpendicular legs $BC=9$ and $CD=4$.
- Its area is half the product of the legs.
- Compared to the whole, that is $18$ out of $96$, or $\tfrac{3}{16}$.
💡 A corner that keeps the original right angle is just a smaller right triangle, so its area is half of leg times leg.
6.RP.A.3 Step 4 Corner at A by a same-height ratio
- Triangle $ABF$ shares the full angle at $A$ with $ACE$.
- Compare in two steps.
- First, $ABF$ and $ACF$ share the same height dropped from $F$ onto line $AC$, so their areas are in the ratio of their bases $AB:AC = 3:12 = \tfrac14$.
- Second, $ACF$ and $ACE$ share the base $AC$, so their areas are in the ratio of $AF:AE = 15:20 = \tfrac34$.
- Multiplying the two shrink factors gives the corner's share of the whole.
💡 When two triangles share an angle, each side you shorten scales the area by that fraction, so you just multiply the two fractions.
6.RP.A.3 Step 5 Corner at E the same way
- Triangle $DEF$ shares the angle at $E$.
- By the identical two-step argument, shortening $EC=16$ down to $ED=12$ scales by $\tfrac{12}{16}=\tfrac34$, and shortening $EA=20$ down to $EF=5$ scales by $\tfrac{5}{20}=\tfrac14$.
- The corner's share is the product.
- Notice all three corners come out equal to $\tfrac{3}{16}$.
💡 The same shared-angle trick works at every corner, and here the fractions just swap places, so the answer repeats.
6.G.A.1 Step 6 Subtract the three corners
- The three corners together take up $\tfrac{3}{16}+\tfrac{3}{16}+\tfrac{3}{16}=\tfrac{9}{16}$ of the whole triangle.
- The inner triangle $DBF$ is everything that remains, so its share is $1-\tfrac{9}{16}=\tfrac{7}{16}$.
- That is choice (E).
💡 Take the whole as $1$, remove the three equal corners, and whatever is left is the middle triangle's share.
8.G.B.6 The triangle is right-angled, but we must find where. The longest side is the hy 6.G.A.1 The three points cut three small triangles off the corners of $ACE$: triangle $A 6.G.A.1 Triangle $BCD$ shares the right angle at $C$, so it is itself a right triangle w 6.RP.A.3 Triangle $ABF$ shares the full angle at $A$ with $ACE$. Compare in two steps. Fi 6.RP.A.3 Triangle $DEF$ shares the angle at $E$. By the identical two-step argument, shor 6.G.A.1 The three corners together take up $\tfrac{3}{16}+\tfrac{3}{16}+\tfrac{3}{16}=\t Review
Reasonableness: The inner triangle clearly sits well inside $ACE$, so its share should be under $\tfrac12$, and $\tfrac{7}{16}$ (just below one half) fits the picture. In real numbers the three corners are $18+18+18=54$, and $96-54=42$, giving $\tfrac{42}{96}=\tfrac{7}{16}$ — the same result. The trap choice $\tfrac{9}{25}$ would come from wrongly using $\left(\tfrac35\right)^2$-type reasoning, and $\tfrac14$ from forgetting the corners are unequal-looking, but the honest corner areas all equal $\tfrac{3}{16}$ and give $\tfrac{7}{16}$.
Alternative: Coordinates confirm it fast. Put $C=(0,0)$, $A=(0,12)$, $E=(16,0)$. Then $B=(0,9)$, $D=(4,0)$, $F=(12,3)$. The shoelace formula gives $[\triangle DBF]=\tfrac12\left|0(0-3)+4(3-9)+12(9-0)\right|=\tfrac12\left|{-24}+108\right|=42$, and $\tfrac{42}{96}=\tfrac{7}{16}$, matching (E).
CCSS standards used (min grade 8)
8.G.B.6Explain a proof of the Pythagorean theorem and its converse (Using $12^2+16^2=20^2$ to locate the right angle at $C$, which fixes the legs and the area of $ACE$.)6.G.A.1Find the area of triangles and other polygons by composing or decomposing into triangles (Splitting $ACE$ into the inner triangle plus three corner triangles, and computing the right-triangle corner and the final subtraction.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Getting each shared-angle corner's area as the product of the two side-length ratios via same-height triangle comparisons.)
⭐ 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.
⭐ 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.
More like this
Same archetype — closest grade level first.
