+0

# Help would be appreciated!

0
83
1

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?

Mar 18, 2023

#1
0

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

Mar 18, 2023