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2012 AMC 10B Problem 12

2012 AMC 10B Problem 12AMC 10 Step by Step

Name: Date: September 13, 2026

Triangles: Area & Pythagorean · difficulty 2 of 5 · about 3 min · position P11-15

Point $B$ is due east of point $A$ . Point $C$ is due north of point $B$ . The distance between points $A$ and $C$ is $10\sqrt 2$ , and $\angle BAC = 45^\circ$ . Point $D$ is $20$ meters due north of point $C$ . The distance $AD$ is between which two integers?

A)
$30 \text{and} 31$
B)
$31 \text{and} 32$
C)
$32 \text{and} 33$
D)
$33 \text{and} 34$
E)
$34 \text{and} 35$

Answer

Key insight

The 45-degree angle makes ABC an isosceles right triangle with legs 10, so AD is the hypotenuse of a 10-by-30 right triangle, sqrt(1000).

Solution

Since $B$ is east of $A$ and $C$ is north of $B$ , the angle at $B$ is a right angle. With $\angle BAC = 45^\circ$ , triangle $ABC$ is a $45$ - $45$ - $90$ triangle whose hypotenuse is $AC = 10\sqrt 2$ , so both legs are $AB = BC = 10$ .

$D$ lies $20$ farther north of $C$ along the same vertical line through $B$ , so $BD = 10 + 20 = 30$ . Triangle $ABD$ is right-angled at $B$ with legs $10$ and $30$ :
$AD = \sqrt{10^2 + 30^2} = \sqrt{1000}.$
Since $31^2 = 961 < 1000 < 1024 = 32^2$ , the distance $AD$ is between $31$ and $32$ .

The answer is $\boxed{\textbf{(B)}\ 31\ \text{and}\ 32}$ .

Why this works

Compass directions ("due east," "due north") announce right angles, so the whole figure is two right triangles sharing the leg $AB$ . The $45^\circ$ angle converts the hypotenuse $10\sqrt2$ into legs of $10$ without any trigonometry. Bracketing a square root between consecutive perfect squares is faster and safer than estimating decimals.

The trap

Using 10 sqrt 2 as a leg instead of the hypotenuse, or misplacing sqrt(1000) because 31^2 = 961 and 32^2 = 1024 were not checked.

Common mistakes

Using 10 sqrt 2 as a leg instead of the hypotenuse, or misplacing sqrt(1000) because 31^2 = 961 and 32^2 = 1024 were not checked.
Computing $CD = 20$ and $AC = 10\sqrt2$ and adding them ( $\approx 34.1$ ) as if $A$ , $C$ , $D$ were collinear.

Techniques

Bound the quantity above/below or estimate to pin it down · Set up the equation/formula and compute; no special trick needed