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 Trisected Triangle (Posted on 2006-02-03)
A right triangle PQR, has its hypotenuse, PR, trisected at points A and B. Two lines, QA and QB are then drawn and k is such that QA^2 + QB^2 = (PR^2) * k. Find the value of k.

 No Solution Yet Submitted by Chris, PhD Rating: 4.5000 (4 votes)

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 Solution | Comment 2 of 12 |
(PR)^2 is equal to (QP)^2+(QR)^2.

1. (QA)^2 = ((2/3)(QP))^2 + ((1/3)(QR))^2

2. (QB)^2 = ((1/3)(QP))^2 + ((2/3)(QR))^2

3. (QA)^2 + (QB)^2 = 5/9 (QP)^2 + 5/9 (QR)^2

4. (QA)^2 + (QB)^2 = k*(PR)^2

5. (QA)^2 + (QB)^2 = k * ((QP)^2 + (QR)^2)

If you distribute the k through, you get:

6. (QA)^2 + (QB)^2 = k*(QP)^2 + k*(QR)^2

7. From line 3, since both coefficients are equal, k = 5/9.

So the answer is, k = 5/9.

Another method is to realize that QP and QR will each be multiplied by 2/3 once, and 1/3 once. Since everything is squared, (2/3)^2 + (1/3)^2 = 4/9 + 1/9 = 5/9. Therefore, using either method, k = 5/9, or I'm an idiot and did something really easy, wrong.

Edited on February 3, 2006, 5:23 pm
 Posted by Justin on 2006-02-03 17:20:02

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