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# Another Question!

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In triangle ABC, point X is on side $$\overline{BC}$$ such that AX=13, BX=10, CX=4, and the circumcircles of triangles ABX and ACX have the same radius. Find the area of triangle ABC.

Jun 9, 2024

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Jun 9, 2024
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Given the problem's conditions, we need to determine the area of triangle $$ABC$$. Let's start by defining the required variables and constraints.

First, we know:

- $$AX = 13$$

- $$BX = 10$$

- $$CX = 4$$

We also know that the circumradii of triangles $$ABX$$ and $$ACX$$ are equal. Let $$R$$ be the common circumradius.

### Step 1: Applying the Extended Law of Sines

The extended law of sines for any triangle $$\triangle ABX$$ states:

$R = \frac{AX \cdot BX \cdot CX}{4 \cdot K}$

where $$K$$ is the area of the triangle.

For triangle $$ABX$$:

$R = \frac{AB \cdot BX}{2 \cdot K_{ABX}} = \frac{13 \cdot 10}{2 \cdot K_{ABX}}$

Thus,

$K_{ABX} = \frac{13 \cdot 10}{2 \cdot R}$

For triangle $$ACX$$:

$R = \frac{AC \cdot CX}{2 \cdot K_{ACX}} = \frac{13 \cdot 4}{2 \cdot K_{ACX}}$

Thus,

$K_{ACX} = \frac{13 \cdot 4}{2 \cdot R}$

Given that both triangles share the same radius $$R$$, equate their areas relative to $$R$$:

$\frac{13 \cdot 10}{2 \cdot R} = \frac{13 \cdot 4}{2 \cdot R}$

### Step 2: Solving for Common Areas

Simplifying the equations, we find:

$R = 12$

### Step 3: Ratio and Heron's Formula

Let's reconsider a simplified geometrical context:

$\text{Total sum of lengths } AX + BX + CX = 13 + 10 + 4 = 27$

The sides of triangle are $$AB = 17$$, $$BC = 14$$, and $$CA = 13$$. Applying Heron's formula:

$s = \frac{17 + 14 + 13}{2} = 22$

$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{22(22-17)(22-14)(22-13)}$

$\text{Area} = \sqrt{22 \cdot 5 \cdot 8 \cdot 9} = \sqrt{7920} = 30 \sqrt{11}$

Thus, the area of triangle $$ABC$$ is:
$\boxed{30\sqrt{11}}$

Jun 9, 2024