Suppose TIP and TOP are isosceles triangles. Also suppose that TI=5, PI=7 and PO=11 .
What is the sum of all possible lengths for TO?
Since TIP is an isosceles triangle, we have TP=TI=5. Let x be the length of the segment OP. Then, we have PO=11=x+5, and PI=7.
We can use the Law of Cosines to find the possible lengths of TO, which is the third side of triangle TOP.
In triangle TPO, we have:
TP^2 + PO^2 - 2(TP)(PO)cos(TPO) = TO^2
Substituting the given values, we get:
5^2 + (x+5)^2 - 2(5)(x+5)cos(TPO) = TO^2
Simplifying, we have:
x^2 + 10x - 25cos(TPO) = 0
Using the quadratic formula, we get:
x = [-10 ± sqrt(100 + 100cos(TPO))]/2
Since x is the length of a line segment, it must be non-negative. Therefore, we have:
-5 ≤ -10 + sqrt(100 + 100cos(TPO))/2 ≤ 0
Solving for cos(TPO), we get:
-1 ≤ cos(TPO) ≤ 0.75
Therefore, the possible values of TPO are:
arccos(0.75) ≤ TPO ≤ π
Using the Law of Cosines again in triangle TOP, we have:
TP^2 + TO^2 - 2(TP)(TO)cos(TPO) = OP^2
Substituting the given values, we get:
5^2 + TO^2 - 2(5)(TO)cos(TPO) = (x+5)^2
Substituting the upper bound for cos(TPO), we get:
5^2 + TO^2 - 2(5)(TO)(0.75) = (x+5)^2
Simplifying, we get:
TO^2 - 7.5TO + 5.75 = 0
Using the quadratic formula, we get:
TO = 5/2 ± sqrt(35)/2
Therefore, the sum of all possible lengths for TO is:
TO1 + TO2 = (5/2 + sqrt(35)/2) + (5/2 - sqrt(35)/2) = 5
