AMC 10 · 2005 · #10

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length

isosceles-trianglepythagorean-theorem identify-subproblemsconvert-to-algebra ↑ Prerequisites: pythagorean-theoremisosceles-triangle

📏 Medium solution 💡 2 insights

Problem

Triangle ABC has AC=BC=7 and AB=2, so it is isosceles with the two equal sides meeting at C. A point D sits on the line through A and B, positioned so that B is between A and D, and the distance CD=8. Find the length BD.

Pick an answer.

(A)

(B)

$2\sqrt{3}$

(C)

(D)

(E)

$4\sqrt{2}$

AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 (Draw a Diagram) turns the words into a picture: an isosceles triangle standing on base AB with D marked farther along the line. The key move the picture reveals is dropping a perpendicular from the apex C straight down to the line, which splits everything into right triangles. Tool #7 (Identify Subproblems) then handles two right triangles in turn — one to find the height, one that contains CD. Tool #4 (Introduce a Variable) names BD so the second right triangle becomes an equation we can solve.

1STEP 1

Drop a perpendicular from C

Drop the height from C to line AB, meeting it at M. Since AC=BC, M is the midpoint, so AM=MB=1; D lies farther out past B.

AM=MB=1/2(2)=1

2STEP 2

Find the height CM

Right triangle CMB has leg MB=1 and hypotenuse CB=7, so $CM^2 = 7^2 - 1^2 = 48$ and the height is CM= $4\sqrt{3}$ .

CM² = 7² - 1² = 48 → CM = 4√(3)

3STEP 3

Set up the right triangle with CD

Let BD=x, so MD = MB + BD = 1+x. Right triangle CMD with hypotenuse CD=8 then gives $48 + (1+x)^2 = 64$ .

48 + (1+x)² = 64

4STEP 4

Solve for BD

Subtracting gives $(1+x)^2 = 16$ , and a length is positive, so $1+x = 4$ and BD = 3 — choice (A).

(1+x)² = 16 → 1+x = 4 → x = 3 → (A)

Answer

Plug back in: with BD=3, the far leg is MD = 1+3 = 4, and the height is CM=4√(3)≈ 6.93. Then CD = √(4² + (4√(3))²) = √(16+48) = √(64) = 8, exactly the given value. The answer is also sensible in size: CD=8 is a bit longer than the height 4√(3)≈ 6.93, so D must sit a little way past M — a leg of 4 fits. The trap choice (D) 5 is the length AD = AB + BD = 2+3, the distance from A rather than from B; the question asks for BD.

💡Key takeaway

Drop a straight height from the tip of an isosceles triangle to split it into right triangles, then let the Pythagorean theorem turn the known lengths into an equation for the one you want.

Drop a perpendicular from C
Find the height CM
Set up the right triangle with CD
Solve for BD

Drop a perpendicular from C

Find the height CM

Set up the right triangle with CD

Solve for BD

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