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Topic: Triangle, Heights' Feet, Eotvos (Read 1709 times) |
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rloginunix
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Triangle, Heights' Feet, Eotvos
« on: Sep 7th, 2015, 9:54am » |
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Triangle Construction via its Heights' Feet, from (Hungarian) Eotvos Competitions
Year 1896, problem 3:
1) Construct a triangle given the feet of its altitudes.
2) Express the lengths of the sides of the solution triangle Y in terms of the lengths of the sides of the triangle X whose vertices are the feet of the altitudes of triangle Y.
(Definition: a foot of an altitude/height is a point of intersection of an altitude with the base this altitude falls on)
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dudiobugtron
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Re: Triangle, Heights' Feet, Eotvos
« Reply #1 on: Aug 17th, 2016, 7:43pm » |
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I looked up some background info for this question, and discovered a theorem which says that:
The altitudes of a triangle (ie: Y) are angle bisectors of the angles of its orthic triangle (ie: X).
This knowledge would allow me to construct the altitudes, by constructing angle bisectors of the orthic triangle, from which I could then construct the triangle - since its sides are perpendicular to the altitudes, and pass through the feet.
Is this theorem expected knowledge to solve the problem? Or is working it out supposed to be part of the puzzle?
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rloginunix
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Re: Triangle, Heights' Feet, Eotvos
« Reply #2 on: Aug 18th, 2016, 8:25am » |
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on Aug 17th, 2016, 7:43pm, dudiobugtron wrote:| Is this theorem expected knowledge to solve the problem? Or is working it out supposed to be part of the puzzle? |
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Difficult to say with absolute certainty - different people take on a given problem possessing different amounts of preexisting knowledge. Some - studied, remember and can connect the above theorem to this problem's solution. Some - studied, remember but fail to connect, some - studied but forgot, etc.
On the other hand, as towr noted, in math (and physics) it is not as much what you know but how well you use what you know - nothing precludes a problem solver from discovering the above fact on his/her own (by playing with Cinderella or GeoGebra).
(in my time, on a verbal exam, I would definitely be asked to prove it - some beauty and elegance is to be found in that proof. And I would be asked if the suggested solution is unique.)
Very well.
Since the first part of the problem is solved, you may, but don't have to, give some thought to the second, computational, part. It is a bit involved and a slightly different construction is helpful.
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dudiobugtron
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Re: Triangle, Heights' Feet, Eotvos
« Reply #3 on: Aug 18th, 2016, 2:36pm » |
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Initial thought: For an equilateral triangle, the sides of X are half the length of the sides of Y. The general formula must be robust enough to include this particular triangle. I will try out some other triangles to see if a general formula 'pops out', and if it does, I will then go about trying to prove it 
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EDIT: It would also need to work for an isoceles triangle in the case where the two long sides get aribtrarily longer and longer. In that case, one of the sides of X approaches (from below) the length of the short side of Y, and the other sides of X approach (from above) half that length.
And also it would have to work for the case where one of the angles gets closer and closer to 90 degrees. In that case, two of the sides of X would approach the same length as each other, and the third side would approach 0.
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| « Last Edit: Aug 18th, 2016, 6:25pm by dudiobugtron » |
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rloginunix
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Re: Triangle, Heights' Feet, Eotvos
« Reply #4 on: Aug 20th, 2016, 11:41am » |
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Without inhibiting the flow of one's creative juices, a hint: it pays to go back and work on a proof of the "orthic triangle theorem".
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