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In triangle ABC, point X is on side BC  such that AX=5, BX=5, CX=4,   and the circumcircles of triangles ABX and ACX have the same radius. Find the area of triangle ABC.

 Mar 19, 2024
 #1
avatar+128406 
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Circumradius =  product of the sides /  [4 * Area]

Since the circumradius of each triangle is the same, then the areas are the same

So this means that  we can  solve this :    

[ AB * AX * BX[  = [ AC * AX * CX]

[AB * 5 * 5 ]  = [ AC * 5 * 4 ]

25AB = 20AC

AB  = (20 / 25)AC

AB = (4/5)AC

 

-cos ( AXB)  =  cos (AXC)

Law of  Cosines

 

AB^2 =  BX^2 + AX^2  - 2(BX * AX) cos (AXB)

AC^2 = CX^2 + AX^2 - 2(CX * AX)cos (AXC)

 

[(4/5)AC]^2  = 5^2  + 5^2  - 2(25) cos (AXB)

AC^2 = 4^2 + 5^2  - 2(20) (-cos AXB)

 

(16/25)AC^2  = 50 - 50cos (AXB)

AC^2  = 41 + 40cos (AXB)

 

cos (AXB)  = cos(AXB)....so......

 

[ (16/25)AC^2  - 50 ] / [-50 ]   =  [AC^2 - 41] / 40

 

40  [ (16/25)AC^2  - 50 ] = -50 [ AC^2 - 41]

 

(25.6)AC^2  - 2000  = -50AC^2 + 2050

 

(75.6)AC^2 =  4050

 

AC^2  = 4050 / 75.6

 

AC =  sqrt [ 4050 / 75.6 ]  ≈ 7.32

 

AB = (4/5)7.32  =  5.86

 

Semiperimeter  ≈   [ AB + AC +BC ]  /  2 =   [ 5.86 + 7.32 + 9  ]  /   2   = 11.088

 

[ABC]   = sqrt [ 11.088 * ( 11.088 - 5.86) (11.088 - 7.32) (11.088 - 9) ] ≈  21.36

 

 

cool cool cool

 Mar 20, 2024

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