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2005 AMC 10B Problem 10

2005 AMC 10B Problem 10AMC 10 Step by Step

Name: Date: September 13, 2026

Triangles: Area & Pythagorean · difficulty 2 of 5 · about 3 min · position P1-10

In $\triangle ABC$ , we have $AC=BC=7$ and $AB=2$ . Suppose that $D$ is a point on line $AB$ such that $B$ lies between $A$ and $D$ and $CD=8$ . What is $BD$ ?

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

Answer

Key insight

The altitude from C hits the midpoint of AB; its length squared is 48, so the foot is 4 from D by Pythagoras and BD = 4 - 1.

Solution

Let $H$ be the foot of the perpendicular from $C$ to line $AB$ . Because $CA = CB$ , the altitude lands on the midpoint of $AB$ , so $AH = HB = 1$ .

Right triangle $CHB$ gives
$CH^2 = CB^2 - HB^2 = 49 - 1 = 48.$

Now $D$ lies on the same line beyond $B$ , so triangle $CHD$ is also right-angled at $H$ :
$HD^2 = CD^2 - CH^2 = 64 - 48 = 16, \qquad HD = 4.$

Since $B$ is between $H$ and $D$ , $BD = HD - HB = 4 - 1 = 3$ .

The answer is $\boxed{\textbf{(A)}\ 3}$ .

Why this works

When several points sit on one line and their distances to an off-line point are known, drop the perpendicular from that point: every distance becomes a hypotenuse over the same altitude. The isosceles condition tells you exactly where the foot lands, which removes all unknowns except the one you want.

Alternative approach

Stewart's theorem or the law of cosines in $\triangle CBD$ (with $\cos\angle CBD = -\frac{1}{7}$ ) each lead to $BD^2 + 2\,BD - 15 = 0$ , whose positive root is $3$ . The altitude method is faster.

The trap

Forgetting to subtract BH = 1 and answering 4, the distance from the altitude's foot to D.

Common mistakes

Forgetting to subtract BH = 1 and answering 4, the distance from the altitude's foot to D.
Adding instead of subtracting ( $4 + 1 = 5$ ), which corresponds to placing $D$ on the wrong side of $B$ .

Techniques

Add construction lines/points (drop altitudes, extend segments, connect centers) · Set up the equation/formula and compute; no special trick needed