2005 AMC 10B Problem 10

All of $A$, $B$, $D$ lie on one line, and we know how far each is from $C$. A perpendicular from $C$ to that line creates right triangles, where the Pythagorean theorem applies.

1. Where the perpendicular lands. Call the foot of the perpendicular $H$. Since $CA = CB = 7$, the triangle is isosceles, and the perpendicular from $C$ hits the exact middle of $AB$. As $AB = 2$, we get $AH = HB = 1$.
2. Height $CH$. In right triangle $CHB$, the hypotenuse is $CB = 7$ and one leg is $HB = 1$:
$$CH^2 = 7^2 - 1^2 = 49 - 1 = 48.$$
(No need to take the square root; we only need $CH^2$.)
3. Distance $HD$. Point $D$ is on the same line, past $B$. Triangle $CHD$ also has a right angle at $H$, with hypotenuse $CD = 8$:
$$HD^2 = 8^2 - 48 = 64 - 48 = 16, \qquad HD = 4.$$
4. Distance $BD$. Going along the line, $H$ comes first, then $B$, then $D$. So $BD = HD - HB = 4 - 1 = 3$.

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

When several points lie on one line and you know their distances to a point off the line, draw the perpendicular from that point. Every distance becomes the hypotenuse of a right triangle with the same height. The isosceles triangle tells you exactly where the perpendicular lands.

Call $BD = x$ and use the Law of Cosines, if you know it. In triangle $ABC$, the angle at $B$ has cosine $\frac{1}{7}$, so the angle $CBD$ next to it has cosine $-\frac{1}{7}$. In triangle $CBD$: $64 = 49 + x^2 + 2x$, so $x^2 + 2x - 15 = 0$, which factors as $(x + 5)(x - 3) = 0$. The positive answer is $x = 3$.

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 $\boxed{\textbf{(A)}\ 3}$.