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Topic: Minimum Triangle? (Read 613 times) |
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Sir Col
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Minimum Triangle?
« on: Feb 20th, 2005, 6:17am » |
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A unit (radius) circle has centre, C.
P is a remote point such that PA and PB are tangents to the circle.
T lies on the minor arc AB.
The tangent passing through T intersects PA and PB at M and N respectively.
The perimeter of triangle PAB is p1.
The perimeter of triangle PMN is p2.
(1) Given that p1 = p2+1, find the length MN when T lies on PC.
(2) Does this minimum length correspond to the minimum perimeter of triangle PMN?
(3) Prove that T lying on PC represents the minimum length, MN.
(4) What if the restriction p1 = p2+1 is removed?
[e]Edited because I missed a trivial result. By the time I came back to correct it, Barukh had already spotted it.  [/e]
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| « Last Edit: Feb 20th, 2005, 8:16am by Sir Col » |
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Barukh
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Re: Minimum Triangle?
« Reply #1 on: Feb 20th, 2005, 7:17am » |
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on Feb 20th, 2005, 6:17am, Sir Col wrote:
*hard*
Does this minimum length correspond to the minimum perimeter of triangle PMN?
Unless I'm missing the obvious, |
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What is "obvious"?
Quote:| ...the answer to the last question is mind blowing! |
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It is not (mind blowing).
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Noke Lieu
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Re: Minimum Triangle?
« Reply #2 on: Feb 20th, 2005, 11:09pm » |
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Well, without attachments, let's see how well I can describe what I think is right. (doesn't feel right though- can't see a way of discribing the number nicely, which often means it isn't right...)
I have some assumptions that seem to be true (for part 1):
MN is parallel to AB
AM=MT and BN=TN. Very probably, they all equal each other.
That means PAB= PMN+AB, thus AB=1
If AB=1, the triangle ABC is equilateral.
Thus CE (sorry- E is the point where where AB intersects CP) is sin60.
Ah... hang on typing I realise that its better to use Pythag and call it [sqrt](3/4). The same number, just looks nicer.
Thus ET= 1-sin60, or 1-sqrt (3/4)
Drop perpendiculars from AB to M and N. Label where they intersect AB G and H. (I have the one closer to a as g)
Angle AMG is 30 degrees. GM= 1-sin60 thus GA= {(1-sin60)/tan 30} (roughly 0.232, unless my fat head has gotten it wrong)
AG=BH
MN=AB-2AG= (roughly) .536
for when these tags work again...
But like I say, can't think of a nice, pertinent way of discribing that number.
4) well, I immediately though about
a) placing p infinately far from c, making the tagents parallel. Thus pab is 4 units longer, if that still applies. (have no formal schooling about dealing with infinities- have had fun trying to learn though)
b)placing p on the circumfrence. But that proved boring.
c) it seems that my assumptions still hold true.
d) seems like I am missing something (could be 2 and 3...) but seems rather open ended, unless that's what you're asking, general case etc
In which case, I am going home
Anyway, eagerly awaiting to find where I went wrong.
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Sir Col
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Re: Minimum Triangle?
« Reply #3 on: Feb 21st, 2005, 12:36pm » |
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No mistakes and nice work, Noke Lieu!
You can get exact values...
sin60 = sqrt(3)/2 and tan30 = 1/sqrt(3), so GA = (1-sin60)/tan30 = (1-sqrt(3)/2)/(1/sqrt(3)).
Multiply top and bottom by sqrt(3):
GA = (sqrt(3)-3/2) = (2sqrt(3)-3)/2 ~= 0.232
Then MN = AB-2AG = 1-2*(2sqrt(3)-3)/2 = 4-2sqrt(3) ~= 0.536.
If you develop what you suspected about the perimeter of triangle PMN you should be able to answer (2) quite easily.
(3) and (4) are much more difficult, without appealing to intuition and the "obvious" symmetry.
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| « Last Edit: Feb 21st, 2005, 12:37pm by Sir Col » |
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