SOLUTION: In triangle $ABC$, point $X$ is on side $BC$ such that $AX = 13,$ $BX = 13,$ $CX = 5,$ and the circumcircles of triangles $ABX$ and $ACX$ have the same radius. Find the area of tr

Algebra ->  Formulas -> SOLUTION: In triangle $ABC$, point $X$ is on side $BC$ such that $AX = 13,$ $BX = 13,$ $CX = 5,$ and the circumcircles of triangles $ABX$ and $ACX$ have the same radius. Find the area of tr      Log On


   

Question 1207592: In triangle $ABC$, point $X$ is on side $BC$ such that $AX = 13,$ $BX = 13,$ $CX = 5,$ and the circumcircles of triangles $ABX$ and $ACX$ have the same radius. Find the area of triangle $ABC$.
Answer by ikleyn(53250) About Me  (Show Source):
You can put this solution on YOUR website!
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In triangle ABC, point X is on side BC such that AX = 13, BX = 13, CX = 5,
and the circumcircles of triangles ABX and ACX have the same radius. Find the area of triangle ABC.
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        I will solve the problem step by step.
        Make a sketch and follow my reasoning. 


     Step 1.  Triangle BAC is isosceles: angle C is congruent to angle B



Law of sine for triangle AXC, written for angle C, says

    sin%28C%29%2Fabs%28AX%29 = 2R%5BAXC%5D,    (1)

where R%5BAXC%5D is the radius of the circle circumscribed about triangle AXC.



Law of sine for triangle AXB, written for angle B, says

    sin%28B%29%2Fabs%28AX%29 = 2R%5BAXB%5D,    (2)

where R%5BAXB%5D is the radius of the circle circumscribed about triangle AXB.


We are given that  R%5BAXC%5D  is equal to  R%5BAXB%5D.  Therefore, from (1) and (2)

    sin%28C%29%2Fabs%28AX%29 = sin%28B%29%2Fabs%28AX%29.    (3)


It implies that  sin(C) = sin(B).  Since C and B are angles of triangle BAC, it implies
that angle C is congruent to angle B.
Thus, triangle ABC is isosceles, and side AC is congruent to side AB.

Step 1 is complete.



     Step 2.  Triangles BAC and AXB are similar



Indeed, both triangles ABC and AXB are isosceles and have common angle B, which is 
the base angle for both triangles.

Thus step 2 is complete.



     Step 3.  Find the lateral sides AB and AC of triangle BAC



Let  t  be the common length of sides AB and AC of triangle BAC (now unknown).

In triangle BAC we know the length of its base BC = 13 + 5 = 18.

In triangle AXB we know the length of its lateral sides  AX = 13  and  BX = 13.

Since triangles  BAC  and  AXB  are similar,  there is  a proportion for ratios of corresponding sides

    abs%28BC%29%2Fabs%28AX%29 = abs%28AB%29%2Fabs%28AX%29   (it is base : lateral side = base : lateral side).


Substituting the values, this proportion takes the form

    18%2Ft = t%2F13.   (4)


From this proportion,  t^2 = 18*13;  hence,  t = sqrt%2818%2A13%29.


Thus, we found out the lateral sides of triangle BAC: |AB| = |AC| = sqrt%2818%2A13%29.


Step 3 is complete.



     Step 4.  Find the area of triangle BAC



For now, we know that triangle BAC is isosceles: AB= AC = sqrt%2818%2A13%29
and its base BC has the length of 13+5 = 18 units.


Draw perpendicular AD (the altitude) from vertex A to the base BC.
It bisects the base BC in two parts of the length  18%2F2 = 9 units.

The length h of the altitude AD will be (apply the Pythagoras)

    h^2 = %28sqrt%2818%2A13%29%29%5E2 - 9%5E2 = 18*13 - 81 = 153;  h = sqrt%28153%29.


Now the area of triangle BAC is

    area%5BBAC%5D = %281%2F2%29%2Aabs%28BC%29%2Ah = %281%2F2%29%2A18%2Asqrt%28153%29 = 9%2Asqrt%28153%29.


At this point, the problem is solved in full.


ANSWER.  The area of triangle BAC is  9%2Asqrt%28153%29,  or about  111.3238519 square units (approximately).

Solved.

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Surely, this problem is of great Math Olympiad level.