2005 AMC 10B Problem 10

Below is the professionally curated solution for Problem 10 of the 2005 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 10B solutions, or check the answer key.

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Concepts:isosceles triangle altitude Pythagorean Theorem

Difficulty rating: 1370

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 $BD?$

$3$

$2\sqrt{3}$

$4$

$5$

$4\sqrt{2}$

Solution:

Let $H$ be the foot of the altitude from $C$ to line $AB.$ Since $AC = BC,$ $H$ is the midpoint of $AB,$ so $BH = 1$ and $CH^2 = 7^2 - 1^2 = 48.$

Applying the Pythagorean theorem in $\triangle CHD,$ where $HD = BH + BD = 1 + BD,$ gives $8^2 = 48 + (1 + BD)^2,$ so $(1 + BD)^2 = 16.$

Then $1 + BD = 4,$ so $BD = 3.$

Thus, A is the correct answer.

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