AMC 10 · 2019 · #16

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference

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

📏 Long solution 💡 4 insights

Problem

In $\triangle ABC$ with a right angle at $C$ , point $D$ lies in the interior of $\overline{AB}$ and point $E$ lies in the interior of $\overline{BC}$ so that $AC=CD,$ $DE=EB,$ and the ratio $AC:DE=4:3$ . What is the ratio $AD:DB?$

$\textbf{(A) }2:3\qquad\textbf{(B) }2:\sqrt{5}\qquad\textbf{(C) }1:1\qquad\textbf{(D) }3:\sqrt{5}\qquad\textbf{(E) }3:2$

Pick an answer.

(A)

2:3

(B)

$2:\sqrt{5}$

(C)

1:1

(D)

$3:\sqrt{5}$

(E)

3:2

View mode:

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

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

#9 Solve an Easier Related Problem 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.

$$AC = CD = 4,\;\; DE = EB = 3$$

💡 Using numbers $4$ and $3$ instead of the ratio lets us compute lengths directly.

#7 Identify Subproblems 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 $\angle CDE = 180^\circ - 90^\circ = 90^\circ$.

$$\angle CDE = 90^\circ$$

💡 Two isosceles triangles share angles with the original right triangle — the leftover angle at $D$ must be the right angle.

#7 Identify Subproblems 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).

$$CE = \sqrt{4^2 + 3^2} = 5$$

💡 A $3$–$4$–$5$ right triangle pops out — the cleanest Pythagorean triple.

#7 Identify Subproblems 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 $AB = \sqrt{AC^2 + BC^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}$.

$$BC = 8,\;\; AB = 4\sqrt{5}$$

💡 Now the big right triangle has all sides known.

#1 Draw a Diagram 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 $AD = 8/\sqrt{5}$.

$$AD = 2 \cdot AC \cdot \dfrac{AC}{AB} = \dfrac{2 \cdot 16}{4\sqrt{5}} = \dfrac{8}{\sqrt{5}}$$

💡 In an isosceles triangle the altitude from the apex bisects the base — a quick way to read $AD$ off the picture.

#1 Draw a Diagram 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 $DB = 2 \cdot 3 \cdot (2/\sqrt{5}) = 12/\sqrt{5}$.

$$DB = \dfrac{2 \cdot 3 \cdot 2}{\sqrt{5}} = \dfrac{12}{\sqrt{5}}$$

💡 Same trick mirrored: isosceles $\triangle DEB$ drops a clean altitude at the midpoint of $\overline{DB}$.

#1 Draw a Diagram 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).

$$AD : DB = 8 : 12 = \boxed{2 : 3}$$

💡 Both sides have the same $\sqrt{5}$ denominator — it cancels in the ratio.

[1] #9 6.RP.A.3 Pick concrete sizes: let $AC = 4$ and $DE = 3$ (the ratio $4 : 3$). Then $CD = A

[2] #7 8.G.A.5 Find the key angle at $D$. Since $\triangle ACD$ is isosceles with $AC = CD$, $\

[3] #7 8.G.B.7 Now $\triangle CDE$ is right-angled at $D$ with legs $CD = 4$ and $DE = 3$. The

[4] #7 8.G.B.7 Compute $BC$ and $AB$. Since $E$ is on $\overline{BC}$ between $B$ and $C$, $BC

[5] #1 7.G.B.4 Find $AD$. In isosceles $\triangle ACD$, drop the altitude from $C$ to $\overlin

[6] #1 7.G.B.4 Find $DB$ the same way. In isosceles $\triangle DEB$ with $DE = EB = 3$, drop th

[7] #1 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.1 Understand the concept of a ratio and use ratio language (Forming and simplifying the final ratio $AD : DB = 8 : 12 = 2 : 3$.)
6.RP.A.3 Use 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.4 Know 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.5 Use 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.7 Apply 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 $\textbf{(A)}$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 8)

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