AMC 10 · 2010 · #14

Grade 8 geometry-2d

similar-trianglesequilateral-trianglethirty-sixty-ninety-triangleangle-sum-triangle identify-subproblems ↑ Prerequisites: angle-sum-triangle

📏 Long solution 💡 3 insights

Problem

Triangle $ABC$ has $AB=2 \cdot AC$ . Let $D$ and $E$ be on $\overline{AB}$ and $\overline{BC}$ , respectively, such that $\angle BAE = \angle ACD$ . Let $F$ be the intersection of segments $AE$ and $CD$ , and suppose that $\triangle CFE$ is equilateral. What is $\angle ACB$ ?

Pick an answer.

(A)

$60^\circ$

(B)

$75^\circ$

(C)

$90^\circ$

(D)

$105^\circ$

(E)

$120^\circ$

AMC 10 2010 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.

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Toolkit + CCSS Solution

Understand

Restated: In triangle ABC the side AB is twice as long as AC. Point D sits on AB and point E sits on BC so that angle BAE equals angle ACD. The segments AE and CD cross at F, and the little triangle CFE turns out to be equilateral. Find the measure of angle ACB.

Givens: $AB = 2 \cdot AC$; $D$ is on $\overline{AB}$ and $E$ is on $\overline{BC}$; $\angle BAE = \angle ACD$; $F$ is the intersection of $AE$ and $CD$; $\triangle CFE$ is equilateral, so each of its angles is $60^\circ$

Unknowns: the measure of $\angle ACB$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #4 Introduce a Variable, #7 Identify Subproblems

The problem is packed with points and equal angles, so the first move is a clean diagram to see how everything connects at F. Then name the shared angle with a variable and chase angles piece by piece: the equilateral triangle pins the angle at F, one small triangle fixes angle A, and the length rule AB = 2*AC finishes the job.

Execute — Answer: C

#1 Draw a Diagram 7.G.A.2 Step 1

Draw and label the figure

Sketch triangle ABC with AB the long side (twice AC).
Mark D on AB, E on BC, and F where AE meets CD.
Shade triangle CFE and mark its three angles as 60 degrees.
Give the shared angle a name: let x be the equal pair of angles.

$$\text{Let } x = \angle BAE = \angle ACD.$$

💡 A picture with every angle labeled turns a wall of words into something you can point at.

#4 Introduce a Variable 7.G.B.5 Step 2

Angle at F is 120 degrees

Points A, F, E lie on one straight line, so the two angles on one side at F must add to 180 degrees.
One of them, angle CFE, is 60 degrees because triangle CFE is equilateral.
The other one, angle CFA, is what is left.

$$\angle CFA = 180^\circ - \angle CFE = 180^\circ - 60^\circ = 120^\circ.$$

💡 A straight line is a 180 degree angle, so a known piece hands you the rest for free.

#7 Identify Subproblems 8.G.A.5 Step 3

Triangle ACF fixes angle FAC

Look at triangle ACF by itself.
Its angle at C is angle ACD = x (since F is on CD), and its angle at F is 120 degrees.
The three angles add to 180, so the angle at A is whatever remains.

$$\angle FAC = 180^\circ - 120^\circ - x = 60^\circ - x.$$

💡 Once two angles of a triangle are known, the third has nowhere to hide.

#7 Identify Subproblems 7.G.B.5 Step 4

Angle BAC comes out to 60 degrees

Because F is on segment AE, angle FAC is the same as angle EAC.
The full corner angle at A splits into angle BAE and angle EAC, and the x's cancel.

$$\angle BAC = \angle BAE + \angle EAC = x + (60^\circ - x) = 60^\circ.$$

💡 The unknown x cancels, so it never mattered what its value was.

#1 Draw a Diagram 7.G.A.2 Step 5

Build a helper point M

Let M be the midpoint of AB.
Then AM is half of AB, and since AB = 2*AC, we get AM = AC.
Triangle AMC now has two equal sides AM and AC with a 60 degree angle between them, which forces all three angles to be 60 degrees.
So triangle AMC is equilateral and MC = AM = MB.

$$AM = \tfrac{1}{2}AB = AC, \quad \angle A = 60^\circ \;\Rightarrow\; \triangle AMC \text{ equilateral}, \quad MA = MB = MC.$$

💡 An isosceles triangle with a 60 degree tip must be fully equilateral.

#7 Identify Subproblems 8.G.A.5 Step 6

Split angle ACB with M

Since MB = MC, triangle MBC is isosceles, so its base angles are equal: angle MCB = angle MBC = angle B.
From the equilateral triangle, angle MCA = 60 degrees.
The whole angle ACB is these two pieces stacked together.

$$\angle ACB = \angle MCA + \angle MCB = 60^\circ + \angle B.$$

💡 Equal sides sit across from equal angles, so MC = MB copies angle B onto the other side.

#7 Identify Subproblems 8.G.A.5 Step 7

Finish with the triangle angle sum

Add the three angles of triangle ABC: angle A is 60, angle ACB is 60 + angle B, and angle B is itself.
Setting the total to 180 solves for angle B, then for angle ACB.

$$60^\circ + \angle B + (60^\circ + \angle B) = 180^\circ \;\Rightarrow\; \angle B = 30^\circ, \quad \angle ACB = 60^\circ + 30^\circ = 90^\circ. \;\; \textbf{(C)}$$

💡 One equation from the angle sum turns everything into a single unknown you can solve.

[1] #1 7.G.A.2 Sketch triangle ABC with AB the long side (twice AC). Mark D on AB, E on BC, and

[2] #4 7.G.B.5 Points A, F, E lie on one straight line, so the two angles on one side at F must

[3] #7 8.G.A.5 Look at triangle ACF by itself. Its angle at C is angle ACD = x (since F is on C

[4] #7 7.G.B.5 Because F is on segment AE, angle FAC is the same as angle EAC. The full corner

[5] #1 7.G.A.2 Let M be the midpoint of AB. Then AM is half of AB, and since AB = 2*AC, we get

[6] #7 8.G.A.5 Since MB = MC, triangle MBC is isosceles, so its base angles are equal: angle MC

[7] #7 8.G.A.5 Add the three angles of triangle ABC: angle A is 60, angle ACB is 60 + angle B,

Review

Reasonableness: Check the triangle we found: angles 60, 30, 90 with the right angle at C. Then AB is the hypotenuse and AC is the side opposite the 30 degree angle, which is always half the hypotenuse, so AB = 2*AC exactly as required. Everything fits, so angle ACB = 90 degrees is consistent.

Alternative: Instead of the midpoint helper, keep angle A = 60 and use the Law of Sines with AB = 2*AC: since AB/sin C = AC/sin B, the length rule gives sin C = 2 sin B, and with B + C = 120 degrees this forces cos C = 0, so C = 90 degrees. Same answer by trigonometry rather than by construction.

CCSS standards used (min grade 8)

7.G.A.2 Draw geometric shapes with given conditions including triangles (drawing the labeled figure and adding the midpoint helper point M)
7.G.B.5 Use facts about supplementary, complementary, vertical, and adjacent angles (getting angle CFA = 120 from the straight line and adding angle pieces at A)
8.G.A.5 Use informal arguments to establish facts about angle sum and exterior angles (the triangle angle sums in ACF and ABC and the isosceles base-angle reasoning)

⭐ Name the mystery angle, chase the angles around the equilateral piece until it cancels, then let the side rule AB = 2*AC snap the last angle into place.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Draw and label the figure

Angle at F is 120 degrees

Triangle ACF fixes angle FAC

Angle BAC comes out to 60 degrees

Build a helper point M

Split angle ACB with M

Finish with the triangle angle sum

Review

CCSS standards used (min grade 8)

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