AMC 10 · 2019 · #16
Grade 8 geometry-2dProblem
In △ABC with a right angle at C, point D lies in the interior of AB and point E lies in the interior of BC so that AC=CD, DE=EB, and the ratio AC:DE=4:3. What is the ratio AD:DB?
Pick an answer.
AMC 10 2019 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 $ABC$ with the right angle at $C$, point $D$ is on segment $\overline{AB}$ and point $E$ is on segment $\overline{BC}$. We are told $AC = CD$, $DE = EB$, and $AC : DE = 4 : 3$. Find the ratio $AD : DB$.
Givens: $\angle C = 90^\circ$ in $\triangle ABC$; $D$ lies on $\overline{AB}$, $E$ lies on $\overline{BC}$; $AC = CD$ (so $\triangle ACD$ is isosceles); $DE = EB$ (so $\triangle DEB$ is isosceles); $AC : DE = 4 : 3$; Choices: (A) $2:3$, (B) $2:\sqrt{5}$, (C) $1:1$, (D) $3:\sqrt{5}$, (E) $3:2$
Unknowns: The ratio $AD : DB$
Understand
Restated: In right triangle $ABC$ with the right angle at $C$, point $D$ is on segment $\overline{AB}$ and point $E$ is on segment $\overline{BC}$. We are told $AC = CD$, $DE = EB$, and $AC : DE = 4 : 3$. Find the ratio $AD : DB$.
Givens: $\angle C = 90^\circ$ in $\triangle ABC$; $D$ lies on $\overline{AB}$, $E$ lies on $\overline{BC}$; $AC = CD$ (so $\triangle ACD$ is isosceles); $DE = EB$ (so $\triangle DEB$ is isosceles); $AC : DE = 4 : 3$; Choices: (A) $2:3$, (B) $2:\sqrt{5}$, (C) $1:1$, (D) $3:\sqrt{5}$, (E) $3:2$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #9 Solve an Easier Related Problem
Tool #1 (Diagram): the problem mixes angles, equal sides, and a point on $\overline{AB}$ — 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 $\overline{AB}$ 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.
Execute — Answer: A
6.RP.A.3 Step 1 - Pick concrete sizes: let $AC = 4$ and $DE = 3$ (the ratio $4 : 3$).
- Then $CD = AC = 4$ and $EB = DE = 3$.
- Set up coordinates with $C = (0,0)$, $A = (0, 4)$, and $B$ on the positive $x$-axis.
💡 Using numbers $4$ and $3$ instead of the ratio lets us compute lengths directly.
8.G.A.5 Step 2 - Find the key angle at $D$.
- Since $\triangle ACD$ is isosceles with $AC = CD$, $\angle DAC = \angle ADC$.
- Since $\triangle DEB$ is isosceles with $DE = EB$, $\angle EDB = \angle EBD$.
- In right triangle $ABC$, $\angle DAC + \angle EBD = 90^\circ$.
- At point $D$ on line $\overline{AB}$, the three angles $\angle ADC + \angle CDE + \angle EDB = 180^\circ$, so $\angle CDE = 180^\circ - (\angle DAC + \angle EBD) - 0 = 180^\circ - 90^\circ = 90^\circ$...
- wait, more carefully: $\angle ADC = \angle DAC$ and $\angle EDB = \angle EBD$, so $\angle ADC + \angle EDB = \angle DAC + \angle EBD = 90^\circ$.
- Hence $\angle CDE = 180^\circ - 90^\circ = 90^\circ$.
💡 Two isosceles triangles share angles with the original right triangle — the leftover angle at $D$ must be the right angle.
8.G.B.7 Step 3 - Now $\triangle CDE$ is right-angled at $D$ with legs $CD = 4$ and $DE = 3$.
- The hypotenuse is $CE = \sqrt{4^2 + 3^2} = 5$ (a $3$–$4$–$5$ triangle).
💡 A $3$–$4$–$5$ right triangle pops out — the cleanest Pythagorean triple.
8.G.B.7 Step 4 - Compute $BC$ and $AB$.
- Since $E$ is on $\overline{BC}$ between $B$ and $C$, $BC = CE + EB = 5 + 3 = 8$.
- Then $AB = \sqrt{AC^2 + BC^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}$.
💡 Now the big right triangle has all sides known.
7.G.B.4 Step 5 - Find $AD$.
- In isosceles $\triangle ACD$, drop the altitude from $C$ to $\overline{AB}$; it bisects $\overline{AD}$ at its midpoint $M$, so $AD = 2 \cdot AM$.
- From right triangle $AMC$ (right angle at $M$), $AM = AC \cdot \cos(\angle CAB)$.
- And in big right triangle $\triangle ABC$, $\cos(\angle CAB) = AC/AB = 4/(4\sqrt{5}) = 1/\sqrt{5}$.
- So $AM = 4 \cdot (1/\sqrt{5}) = 4/\sqrt{5}$ and $AD = 8/\sqrt{5}$.
💡 In an isosceles triangle the altitude from the apex bisects the base — a quick way to read $AD$ off the picture.
7.G.B.4 Step 6 - Find $DB$ the same way.
- In isosceles $\triangle DEB$ with $DE = EB = 3$, drop the altitude from $E$ to $\overline{AB}$; it bisects $\overline{DB}$.
- So $DB = 2 \cdot EB \cdot \cos(\angle EBD)$, and $\cos(\angle EBD) = \cos(\angle ABC) = BC/AB = 8/(4\sqrt{5}) = 2/\sqrt{5}$.
- So $DB = 2 \cdot 3 \cdot (2/\sqrt{5}) = 12/\sqrt{5}$.
💡 Same trick mirrored: isosceles $\triangle DEB$ drops a clean altitude at the midpoint of $\overline{DB}$.
6.RP.A.1 Step 7 - Form the ratio: $AD : DB = \dfrac{8}{\sqrt{5}} : \dfrac{12}{\sqrt{5}} = 8 : 12 = 2 : 3$.
- The answer is choice (A).
💡 Both sides have the same $\sqrt{5}$ denominator — it cancels in the ratio.
6.RP.A.3 Pick concrete sizes: let $AC = 4$ and $DE = 3$ (the ratio $4 : 3$). Then $CD = A 8.G.A.5 Find the key angle at $D$. Since $\triangle ACD$ is isosceles with $AC = CD$, $\ 8.G.B.7 Now $\triangle CDE$ is right-angled at $D$ with legs $CD = 4$ and $DE = 3$. The 8.G.B.7 Compute $BC$ and $AB$. Since $E$ is on $\overline{BC}$ between $B$ and $C$, $BC 7.G.B.4 Find $AD$. In isosceles $\triangle ACD$, drop the altitude from $C$ to $\overlin 7.G.B.4 Find $DB$ the same way. In isosceles $\triangle DEB$ with $DE = EB = 3$, drop th 6.RP.A.1 Form the ratio: $AD : DB = \dfrac{8}{\sqrt{5}} : \dfrac{12}{\sqrt{5}} = 8 : 12 = Review
Reasonableness: Quick sanity: $AD + DB$ should equal $AB$. $AD + DB = 8/\sqrt{5} + 12/\sqrt{5} = 20/\sqrt{5} = 4\sqrt{5}$, which matches $AB = 4\sqrt{5}$. 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 $\overline{AB}$.
Alternative: Tool #13 (Algebra) with coordinates: place $C = (0,0)$, $A = (0,4)$, $B = (8,0)$. The line $\overline{AB}$ has equation $y = 4 - x/2$. Point $D$ on this line with $CD = 4$ satisfies $x^2 + (4-x/2)^2 = 16$, giving $x = 16/5$. Then $AD = \sqrt{(16/5)^2 + (4-8/5-4)^2}$... and $DB$ similarly — same ratio $2:3$ pops out without the isosceles trick.
CCSS standards used (min grade 8)
6.RP.A.1Understand the concept of a ratio and use ratio language (Forming and simplifying the final ratio $AD : DB = 8 : 12 = 2 : 3$.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Replacing the abstract ratio $AC : DE = 4 : 3$ with concrete lengths $4$ and $3$.)7.G.B.4Know the formulas for area and circumference of a circle (Using right-triangle trigonometry (cosine ratio) to compute $AD$ and $DB$ from the altitude foot in each isosceles triangle. (Closest available — the actual idea is basic trig in right triangles.))8.G.A.5Use informal arguments to establish facts about angle sum and exterior angles (Showing $\angle CDE = 90^\circ$ by adding angles on a straight line and using the right triangle's $\angle A + \angle B = 90^\circ$.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Computing $CE = 5$ from $CD = 4, DE = 3$ and $AB = 4\sqrt{5}$ from $AC = 4, BC = 8$.)
⭐ 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 $\textbf{(A)}$.
⭐ 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 $\textbf{(A)}$.
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