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Problem 732

AIME late

Geometry Difficulty 5.4 Find the answer

[Pythagorean Theorem (direct and inverse).] Law of Sines $\quad]$

In triangle $ABC$ , angle $C$ is a right angle, the tangent of angle $A$ is $\frac{\mathbf{1}}{\mathbf{4}}$ , and the median $BD$ is $\sqrt{5}$ . Find the area of triangle $ABD$ and the radius of the circumcircle of triangle $ABD$ .

Your answer

Official solution

Let $B C=x, \angle B A C=\alpha$ . Then

$A C=\frac{B C}{\lg r_{2}}=\frac{\frac{x}{1}}{1}=4 x$

Applying the Pythagorean theorem to triangle $B C D$ , we get

$B D 2=B C 2+C D 2, \text { or } x^2+4x^2=5,$

from which $x=1$ . Since the median of a triangle divides it into two equal-area triangles,

$S_{\triangle A B D}=\frac{1}{2} S_{\triangle A B C}=\frac{1}{2} \cdot \frac{1}{2} B C \cdot A C=\frac{1}{4} \cdot 1 \cdot 4=1 .$

$\cos 2 \alpha=\frac{1}{1+l 2^{2}{ }^{2} 2}=\frac{1}{1+\frac{1}{16}}=\frac{16}{17}, \sin 2 \alpha=\frac{1}{\sqrt{17}}$

If $R$ is the radius of the circumcircle of triangle $A B D$ , then

$R=\frac{B D}{2 \sin \angle B A D}=\frac{\sqrt{5}}{2 \cdot \frac{1}{\sqrt{17}}}=\frac{\sqrt{85}}{2} .$

## Answer

$S_{\triangle A B D}=1 ; R=\frac{\sqrt{85}}{2}$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.