|
|
|
|
|
|
|
|
CTK Exchange
mr_homm
Member since May-22-05
|
Oct-09-07, 06:47 AM (EST) |
    |
2. "RE: An interesting isosceles triangle."
In response to message #0
| |
Yes, this is a beautiful puzzle. Here is my solution: Extend BC and PQ to meet at D. Draw AD to complete triangle ADB. Since PD is the perpendicular bisector of AB, ABD is isoceles, with AD = BD. Since Q is on the bisector PD, it follows by symmetry that BQ = AQ, so that BQA is isoceles. Now let angle BAQ = ABQ = a and angle CBQ = CQB = b (because BC = CQ, so that BCQ is isoceles). Then at angle AQB = pi - 2a, and this angle is supplementary with b. Therefore, pi - 2a + b = pi, so that b = 2a. But since BQ divides angle ABC into a and b, and since ABC is isoceles, the sum of the angles on ABC is a + 2(a+2a) = 7a = pi. Thus a = pi/7. Perhaps there are other methods of solution as well. Thank you for an interesting problem! --Stuart Anderson |
|
|
Alert | IP |
Printer-friendly page |
Reply |
Reply With Quote | Top |
|
|
 |
alexb
Charter Member
2093 posts |
Oct-09-07, 07:00 AM (EST) |
|
|
3. "RE: An interesting isosceles triangle."
In response to message #2
| |
>Perhaps there are other methods of solution as well. Yes, of course. Triangle BCQ is isosceles. If angle BCQ = α, angle CBQ = (π - α)/2. Angle ABC = α, hence angle ABQ = α - CBQ = 3α/2 - π/2. Triangle ABQ is isosceles because PQ is the perpendicualr bisector of AB. Hence angle BAC = ABQ = 3α/2 - π/2. On the other hand, angle BAC = π - 2α. Equating the two expressions gives 7α/2 = 3π/2 so that α = π/7. |
|
|
Alert | IP |
Printer-friendly page |
Reply |
Reply With Quote | Top |
|
|
|
|
mr_homm
Member since May-22-05
|
Oct-09-07, 10:57 AM (EST) |
|
|
4. "RE: An interesting isosceles triangle."
In response to message #3
| |
I see from your proof that my construction of point D was unnecessary. I noticed that ADB was an isoceles triangle and used that to show that AQB was isoceles. However, as you point out, the reasoning I applied to ADB would apply directly to AQB just as well. I think I prefer your method, which has less extraneous construction. Here is a related problem: An isoceles triangle ABC with AB=AC is tiled by two smaller isoceles triangles. What are the possible values for angle A? I believe there are exactly 3 solutions, one of which is of course pi/7. The stipulation that all triangles be isoceles makes the solutions more direct than in the original problem, but adds the complication of cases. --Stuart Anderson |
|
|
Alert | IP |
Printer-friendly page |
Reply |
Reply With Quote | Top |
|
|
|
You may be curious to have a look at the old CTK Exchange archive.
Please do not post there.
![]()
Copyright © 1996-2018 Alexander Bogomolny
|
|
Printer-friendly copy
Email this topic to a friend
