What's the value of x? - August 12-16, 1996
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Both legs of an isosceles triangle are radii of a circle, and the length of each radius is 6. The distance from the center of the circle to a point P on the base of the triangle is 4. If the distances from P to the triangle's other vertices are 5 and x, what is the value of x?
Not a lot of answers received this week, but detailed descriptions in various formats. Justin Lam sent us two possible solutions - see below.Justin Lam Grade: 7th/8th School: Sequoia Middle School Given: Triangle ABC is isosceles with AB = AC = 6 Let P be a point on BC such that AP = 4 and PC = 5 Let x = BP. Find: x Let <ABC = a. Then <ACB = a because Triangle ABC is isosceles. Apply Laws of Cosines on Triangle APC we get 4^2 = 5^2+6^2-2(5)(6)cos(a) 16 = 25+36-60cos(a) cos(a) = 3/4 Apply Laws of Cosines again. This time on Triangle ABP we get 4^2 = 6^2+x^2-2(6)(x)cos(a) 16 = 36+x^2-12xcos(a) 16 = 36+x^2-12x(3/4) 16 = 36+x^2-9x x^2-9x+20 = 0 Solving for x by factoring (x-5)(x-4) = 0 x = 4 or 5 x cannot be 5 because if it is, then BP=PC=5. This would mean that AP is the altitude of the isosceles Triangle ABC. This would also mean that Triangle ABP is a right triangle with sides 4, 5, 6 which contradicts the Pythagorean Theorem. So, x = 4. Here's another possible solution: Given: Isosceles Triangle ABC with AB=AC=6. P is on BC such that AP=4 and PC=5. If BP=x, find x. Let Q be on BC such that AQ is the altitude for Triangle ABC. Let AQ=h. Because Triangle ABC is isosceles, AQ is perpendicular to BC and bisects BC (i.e. BQ=QC=(5+x)/2). Also, PC=5 implies QP=|(5+x)/2-5|=|(5-(5+x)/2)|. Now we have two right triangles with Triangle AQC = AQ--QC--AC = h--(5+x)/2--6 and Triangle AQP = AQ--QP--AP = h--|(5+x)/2-5|--4 . Therefore, 6^2 = h^2+((5+x)/2)^2 and 4^2 = h^2+((5+x)/2-5)^2 by the Pythagorean Theorem. Subtracting these two equations, we get 20 = ((5+x)/2)^2 - ((5+x)/2-5)^2 or 20 = (5+x)/2)^2 - ((5+x)/2)^2 - 10(5+x)/2 + 25) or 20 = 5(5+x) - 25 or 5x = 20 or x = 4.
Thomas S. Kuo School: Murray Middle School, Ridgecrest, California Grade: 8th O * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * A P B O is the center of a circle OA and OB are radii. OA = OB = 6 OP = 4 AP = 5 PB = x x = 4 (1) Use the Law of Cosines 4^2 = 6^2 + 5^2 - 2 * 6 * 5 * cos(A) in triangle OAP 4^2 = 6^2 + x^2 - 2 * 6 * x * cos(B) in triangle OBP (2) Solve the equations for x. (a) 4^2 = 6^2 + 5^2 - 2 * 6 * 5 * cos(A) cos(A) = 45/60 = 3/4 (b) cos(A) = cos(B) = 3/4 (c) 4^2 = 6^2 + x^2 - 2 * 6 * x * cos(B) 4^2 = 6^2 + x^2 - 2 * 6 * x * 3/4 x^2 - 9x + 20 = 0 (x - 4) * (x - 5) = 0 x = 4 or 5 (3) Prove that x = 5 is not valid. If x = 5, then OP is perpendicular to AB because P would then be the midpoint of AB. If you use the Pythagorean Theorem to test if 4^2 + 5^2 = 6^2, the result is that they are not equal.
Chui Yuk Man Grade: Secondary 6 School: Queen Elizabeth School (Hong Kong) Answer: Let the centre of the circle be O and the isosceles triangle be ABC, with A and B both on the circumference of the circle. Point P is on the line AB, and let the mid-point of AB be C. OA^2-AC^2=OC^2 OP^2-PC^2=OC^2 OA^2-AC^2=OP^2-PC^2 36-((x+5)/2)^2=16-((x+5)/2-x)^2 (It doesn't matter whether it is ((x+5)/2-x)^2 or ((x+5)/2-5)^2 since they are the same) Solving the above equation, the value of x is 4.
Brian Gordon Dartmouth '92 The answer is 4. Drop the perpendicular from the center of the circle. It divides the base of the triangle in half, and divides the segment of length 5 into smaller segments of p and 5-p. I have the diagram set up so the "p" is the middle of the three segments. That way there's a right triangle formed by the p, the altitude of the triangle, and the given segment of length 4. To get an expression for the height, use the right triangle with hypotenuse 6 and leg 5-p. The height is then sqrt(36 - (5-p)^2) = sqrt(11+10p-p^2). Use this expression with the first right triangle to get that [sqrt(11+10p-p^2)]^2 + p^2 = 4^211+10p-p^2 + p^2 = 1611+10p = 1610p = 5p=.5 which means 5-p = 4.5, and since the original triangle was isosceles, then x+p = 4.5 so x = 4.
Ken Duisenberg Stanford University '93 Hewlett Packard Engineer Answer: 4. Solution: Draw a line perpendicular to the base that goes through the center of the circle, let the length of this line be 'd'. Then there are two right triangles to which to apply the Pythagorean Theorem. The two options below allow for easy understanding if x is (a) less than or (b) greater than 5, and both lead to the same solution: (1) d^2 + [ (5+x)/2 ]^2 = 6^2 (2a) d^2 + [ ((5+x)/2) - x ]^2 = 4^2, OR (2b) d^2 + [ ((5+x)/2) - 5 ]^2 = 4^2 Subtract (1-2): (3a) [ (5+x)/2 ]^2 - [ ((5+x)/2) - x ]^2 = 20, OR (3b) [ (5+x)/2 ]^2 - [ ((5+x)/2) - 5 ]^2 = 20 Expand the second term and subtract the first: (4a) x(5+x) - x^2 = 20, OR (4b) 5(5+x) - 25 = 20 Expand, subtract, and solve for x: (5a) 5x = 20 --> x = 4, OR (5b) 5x = 20 --> x = 4.
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