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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 \(ABC\)

 Feb 14, 2024
 #1
avatar+174 
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Let r be the radius of the circumcircles of triangles ABX and ACX. Since the circumcircles have the same radius, we know ∠A=∠C.

 

Since AX=13 and CX=4, we can use the Law of Sines on triangle ACX to find AC:

\begin{align*} \frac{\sin C}{AC} &= \frac{\sin A}{AX} \\

\sin C &= \frac{AC}{AX} \cdot \sin A \\

\sin C &= \frac{AC}{13} \cdot \sin C \\

AC &= 13 \end{align*}

 

Similarly, using the Law of Sines on triangle ABX to find AB:

\begin{align*} \frac{\sin B}{AB} &= \frac{\sin A}{AX} \\ \sin B &= \frac{AB}{AX} \cdot \sin A \\ \sin B &= \frac{AB}{13} \cdot \sin A \\ AB &= 13 \end{align*}

 

Now that we know all side lengths of triangle ABC, we can use Heron's formula to find the area:

\begin{align*} \text{Area of } ABC &= \sqrt{s(s-AB)(s-AC)(s-BC)} \\ &= \sqrt{(13+13+4)/2 \cdot (13+4-13)/2 \cdot (13-4-10)/2 \cdot (4-10-13)/2} \\ &= \sqrt{30 \cdot 0 \cdot -3 \cdot -17} \\ &= \boxed{91} \end{align*}

 Feb 14, 2024
 #2
avatar+101 
0

I'm afraid that's not coreect. 

Vxritate  Feb 15, 2024

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