Algebra.Com's Answer #845556 by ikleyn(53298)![\"\"](\"/images/stars/stars-13.gif\")  
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document.write( "In triangle ABC, point X is on side BC such that AX = 13, BX = 13, CX = 5,
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document.write( "and the circumcircles of triangles ABX and ACX have the same radius. Find the area of triangle ABC.
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document.write( " I will solve the problem step by step.\r
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document.write( " Make a sketch and follow my reasoning.\r
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document.write( " Step 1. Triangle BAC is isosceles: angle C is congruent to angle B\r\n" );
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document.write( "Law of sine for triangle AXC, written for angle C, says\r\n" );
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document.write( "  =  , (1)\r\n" );
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document.write( "where  is the radius of the circle circumscribed about triangle AXC.\r\n" );
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document.write( "Law of sine for triangle AXB, written for angle B, says\r\n" );
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document.write( "  =  , (2)\r\n" );
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document.write( "where  is the radius of the circle circumscribed about triangle AXB.\r\n" );
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document.write( "We are given that is equal to . Therefore, from (1) and (2)\r\n" );
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document.write( " = . (3)\r\n" );
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document.write( "It implies that sin(C) = sin(B). Since C and B are angles of triangle BAC, it implies\r\n" );
document.write( "that angle C is congruent to angle B.\r\n" );
document.write( "Thus, triangle ABC is isosceles, and side AC is congruent to side AB.\r\n" );
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document.write( "Step 1 is complete.\r\n" );
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document.write( " Step 2. Triangles BAC and AXB are similar\r\n" );
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document.write( "Indeed, both triangles ABC and AXB are isosceles and have common angle B, which is \r\n" );
document.write( "the base angle for both triangles.\r\n" );
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document.write( "Thus step 2 is complete.\r\n" );
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document.write( " Step 3. Find the lateral sides AB and AC of triangle BAC\r\n" );
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document.write( "Let t be the common length of sides AB and AC of triangle BAC (now unknown).\r\n" );
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document.write( "In triangle BAC we know the length of its base BC = 13 + 5 = 18.\r\n" );
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document.write( "In triangle AXB we know the length of its lateral sides AX = 13 and BX = 13.\r\n" );
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document.write( "Since triangles BAC and AXB are similar, there is a proportion for ratios of corresponding sides\r\n" );
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document.write( "  =  (it is base : lateral side = base : lateral side).\r\n" );
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document.write( "Substituting the values, this proportion takes the form\r\n" );
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document.write( "  =  . (4)\r\n" );
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document.write( "From this proportion, t^2 = 18*13; hence, t =  .\r\n" );
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document.write( "Thus, we found out the lateral sides of triangle BAC: |AB| = |AC| = .\r\n" );
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document.write( "Step 3 is complete.\r\n" );
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document.write( " Step 4. Find the area of triangle BAC\r\n" );
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document.write( "For now, we know that triangle BAC is isosceles: AB= AC = \r\n" );
document.write( "and its base BC has the length of 13+5 = 18 units.\r\n" );
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document.write( "Draw perpendicular AD (the altitude) from vertex A to the base BC.\r\n" );
document.write( "It bisects the base BC in two parts of the length  = 9 units.\r\n" );
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document.write( "The length h of the altitude AD will be (apply the Pythagoras)\r\n" );
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document.write( " h^2 =  -  = 18*13 - 81 = 153; h =  .\r\n" );
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document.write( "Now the area of triangle BAC is\r\n" );
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document.write( "  =  =  =  .\r\n" );
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document.write( "At this point, the problem is solved in full.\r\n" );
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document.write( "ANSWER. The area of triangle BAC is , or about 111.3238519 square units (approximately).\r\n" );
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document.write( "Solved.\r
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document.write( "Surely, this problem is of great Math Olympiad level.\r
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