AMC 10 · 2004 · #24

Grade 11 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

trigonometric-ratiostangent-addition-formulasequences-geometricsequences-arithmetic convert-to-algebraidentify-subproblems ↑ Prerequisites: trigonometric-ratios

📏 Long solution 💡 4 insights 📊 Diagram

Problem

An isosceles triangle has an altitude from its apex, and the base is extended past one end to a point at distance 10 from the apex. Three tangents form a geometric progression and three cotangents form an arithmetic progression. Find the area of the triangle.

Pick an answer.

(A)

(B)

$\frac {50}3$

(C)

$10\sqrt{3}$

(D)

$8\sqrt{5}$

(E)

How to solve

Strategy Introduce a Variable

Every angle in the problem has its vertex at B and one side along the line ADCE, and BD is perpendicular to that line. So each of those angles has a tangent equal to (a length along the line) divided by BD. Naming three lengths — the altitude, the half-base, and the distance from the foot to E — turns both progression conditions into algebra in those three lengths, with no angle measure ever needed. The two conditions then pin down two ratios, and BE = 10 fixes the scale.

1STEP 1

Fix the picture and the foot D

The altitude is also the median, so it splits the apex angle evenly.

AD = DC, ∠ ABD = ∠ DBC, BD ⊥ AE

2STEP 2

Name three lengths

Naming the height and two horizontal pieces gives every tangent directly.

h = BD, p = DA = DC, q = DE, tan ∠ DBC = p/h, tan ∠ DBE = q/h

3STEP 3

Express the outer two tangents

Sum and difference formulas write the outer two tangents as clean fractions.

tan ∠ CBE = (q/h - p/h)/(1 + pq/h²) = (h(q-p))/(h² + pq), tan ∠ ABE = (p/h + q/h)/(1 - pq/h²) = (h(q+p))/(h² - pq)

4STEP 4

Use the geometric progression

The geometric condition forces the far piece to equal the height.

q²/h² = (h(q-p))/(h²+pq) · (h(q+p))/(h²-pq) = (h²(q²-p²))/(h⁴ - p²q²) ⟹ q²h⁴ - p²q⁴ = q²h⁴ - p²h⁴ ⟹ q = h

5STEP 5

Use the arithmetic progression

The arithmetic condition then makes the height three times the half-base.

2·(h+p)/(h-p) = 1 + h/p = (p+h)/p ⟹ 2/(h-p) = 1/p ⟹ 2p = h - p ⟹ h = 3p

6STEP 6

Bring in BE = 10

The given length of 10 fixes the size through a right triangle.

BE² = h² + q² = 2h² = 100 ⟹ h = 5√(2), p = 5√(2)/3

7STEP 7

Compute the area

Base times height halved gives 50/3, choice (B).

[ABC] = 1/2 · AC · BD = 1/2(2p)h = ph = 5√(2)/3 · 5√(2) = 50/3

Answer

50/3

Rebuild the figure from the answer and test both conditions numerically. With h = 5√(2), p = 5√(2)/3, q = 5√(2), the tangents are tan ∠ DBC = p/h = 1/3, tan ∠ DBE = 1, tan ∠ CBE = (h(q-p))/(h²+pq) = 1/2, and tan ∠ ABE = (h(q+p))/(h²-pq) = 2. So the first progression is 1/2, 1, 2 — geometric with ratio 2 — and the cotangents are 1, 2, 3 — arithmetic with difference 1. Both conditions hold exactly, and BE = √(50+50) = 10 as required. The configuration is also unique: the geometric condition forced q = h and the arithmetic condition forced h = 3p, each with no second solution, so no other triangle competes. Size-wise, AC = 10√2/3 ≈ 4.71 and BD = 5√2 ≈ 7.07 give area ≈ 16.67, a tall narrow triangle — matching cot ∠ DBC = 3, a narrow apex. That 16.67 sits just above choice (A) 16, so the exact fraction, not an estimate, is what decides between them. One caution: in the printed figure E is drawn much closer to C than it truly is, so the picture is a guide to the order of the points, not to their distances.

💡Key takeaway

When every angle in a problem shares a vertex and a perpendicular, replace the angles by lengths along that perpendicular's base line and the trigonometry turns into ordinary algebra.

Fix the picture and the foot D
Name three lengths
Express the outer two tangents
Use the geometric progression
Use the arithmetic progression
Bring in BE = 10
Compute the area