# Thread: Find the area of triangle

1. ## Find the area of triangle

Hello everyone, pls help me with this problem. Thanks in advance!

Three unit circles are arranged side by side in a row. A is the point on the first circle
that is collinear with the centers of the three circles which does not touch the second circle. The lines AB and AC are tangent
to the third circle at the points B and C. Find the area of ABC.

2. ## Re: Find the area of triangle

Let O be the center of the third circle. Triangle ABO is right. From it you can find AB and angle BAO, which is half of angle BAC.

3. ## Re: Find the area of triangle

Hello, shiny718!

Three unit circles are arranged side by side in a row.
A is the point on the first circle that is collinear with the centers of the three circles
. . and does not touch the second circle.
The lines AB and AC are tangent to the third circle at the points B and C.
Find the area of ABC.

Code:
              * * *               * * *           B   * * *
*           *       *           *       o           *
*               *   *               *   *               *
*                 * *                 * *                 *

*                   *                   *                   *
A o         *         *         *         *         o         *
*                   *                   *         P         *

*                 * *                 *                   *
*               *   *               *   *               *
*           *       *           *       o           *
* * *               * * *           C   * * *
$\text{Let }P\text{ = center of the third circle. }\text{ Draw radii }PB, PC.$

$\text{Draw tangents }AB,\,AC,\text{ and line-of-centers }AP.\;\text{ Let }\theta = \angle BAP.$

$\text{In right triangle }ABP\!:\:BP = 1,\:AP = 5.$
. . $\text{Hence: }\,AB \,=\,AC \,=\,\sqrt{24}$

$\text{Note that: }\:\sin\theta \,=\,\frac{BP}{AP} \,=\,\frac{1}{5} \quad\Rightarrow\quad \cos\theta \,=\,\frac{\sqrt{24}}{5}$

$\text{Area of }\Delta ABC \:=\:\tfrac{1}{2}(AB)(AC)\sin2\theta \;=\;\tfrac{1}{2}\left(\sqrt{24}\right)\left(\sqrt {24}\right)\;\!2\sin\theta\cos\theta$

. . . . . . . . . . . . $=\;24\cdot\frac{1}{5}\cdot\frac{\sqrt{24}}{5} \;=\;\frac{48\sqrt{6}}{25}$

Edit: too slow . . . again!

4. ## Re: Find the area of triangle

A geometric solution ( reference Soroban's solution)

Let D = intersection of BC and AO
BO =1 AO =5 AB= rad 24
Triangle ABO is a rt triangle @ B. BD is the altitude on AO
let DO = x AD = 5-x
AB^2 /BO^2 = 5-x/x
24/1 = 5-x /x whence x=1/5 and 5-x= 24/5
BD^2 = 24/5 *1/5