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In triangle ABC, point X is on side 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. (Someone else asked the same question a bit ago, but nobody answered it, so I'm asking on a new strand)

 May 7, 2023
 #1
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The circumcircle of a triangle is the circle that passes through all three vertices of the triangle. The radius of the circumcircle of a triangle is the distance from any vertex of the triangle to the center of the circumcircle.

In triangle ABX, the radius of the circumcircle is AB/2​=13/2​.

In triangle ACX, the radius of the circumcircle is AC/2​=14/2​=7.

Since the circumcircles of triangles ABX and ACX have the same radius, we have AB/2​=AC/2​, or AB=AC.

Since AB=AC, we have △ABC is an isosceles triangle.

The area of an isosceles triangle is 21​⋅b⋅h, where b is the base of the triangle and h is the height of the triangle.

In triangle ABC, the base is BC=10 and the height is AX=13.

Therefore, the area of triangle ABC is 1/2⋅10⋅13=65​.

 May 7, 2023
 #2
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Let r be the radius of the circumcircles of triangles ABX and ACX. Then the area of triangle ABX is 1/2​(AB)(r)=1/2​(13)(r), and the area of triangle ACX is 1/2​(AC)(r)1/2​(14)(r). Since the circumcircles have the same radius, the areas of the triangles are equal. Therefore, 1/2​(13)(r)=1/2​(14)(r), or r=13.

The area of triangle ABC is1/2​(AB)(AC) = 1/2(13)(14) = 91​.

 May 8, 2023

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