AMC 10 · 2005 · #6

Grade 8 geometry-2d
isosceles-trianglepythagorean-theorem identify-subproblemsconvert-to-algebra ↑ Prerequisites: pythagorean-theoremisosceles-triangle
📏 Medium solution 💡 2 insights
Problem
An isosceles triangle has two equal sides of 7 and a base of 2. A point sits on the base line beyond one end, at distance 8 from the apex. Find how far that point is from the near end of the base.

Pick an answer.

(A)
3
(B)
$2\sqrt{3}$
(C)
4
(D)
5
(E)
$4\sqrt{2}$
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

Symmetry puts the foot of the height at the base's midpoint.

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

Find the height CM

One right triangle gives the height.

CM² = 7² - 1² = 48 → CM = 4√(3)
3STEP 3

Set up the right triangle with CD

A second right triangle brings in the given long distance.

48 + (1+x)² = 64
4STEP 4

Solve for BD

Solving gives 3, choice (A).

(1+x)² = 16 → 1+x = 4 → x = 3 → (A)
Answer
3
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