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Topic: Isosceles Triangle (Read 763 times) |
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william wu
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Isosceles Triangle
« on: May 13th, 2003, 8:44am » |
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Via e-mail. A math problem for "gifted high-schoolers":
You are given an isosceles triangle ABC, where |AB| = |BC|. Draw the set of points { M : angle AMB = angle CMB }.
"Hint: there is a grade C answer, a grade B answer, and a grade A answer."
BTW, if anyone has a better name for this puzzle, feel free to suggest.
Update 10:23 AM 5/13/2003: s/isoceles/isosceles
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| « Last Edit: May 13th, 2003, 10:23am by william wu » |
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ThudnBlunder
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Re: Isosceles Triangle
« Reply #1 on: May 13th, 2003, 10:19am » |
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Quote:| BTW, if anyone has a better name for this puzzle, feel free to suggest. |
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How about Isosceles Triangle? 
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Kozo Morimoto
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Re: Isosceles Triangle
« Reply #3 on: May 13th, 2003, 6:12pm » |
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Is it
the perpendicular plane that bisects AC and at the same time go thru B (but avoiding the point B) ? Not sure on the mathematical definition of an angle for AMB where M is B?
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tohuvabohu
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I think that would be the grade C answer. The B answer would be the line that passes through A and C minus the points between A and C. I'm not sure what the A answer would be yet. The next time I meet a gifted high-schooler I'll ask him.
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MattyDK23
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Re: Isosceles Triangle
« Reply #5 on: May 13th, 2003, 10:06pm » |
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What about the answer being S = S1 U S2, where
S1 = {m : m is a point in the perpendicular plane that bisects AC and at the same time go thru B (but avoiding the point B) }
S2 = {m: m is a point in line that passes through A and C, minus the points between A and C }
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Leonid Broukhis
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So far so good, just one little piece left. And as the requirement is to "draw" the set of points, {M} is restricted to the plane ABC.
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James Fingas
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Re: Isosceles Triangle
« Reply #7 on: May 14th, 2003, 9:19am » |
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Here's a hint for the A answer:
Picture a line drawn through B that passes very near A. Call the points on this line N. For points in the vicinity of A, the angle CNB is roughly constant. However, the angle ANB changes from very large to very small. So CNB starts out smaller than ANB, and then becomes larger. Somewhere in the middle, it must be equal...
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Lightboxes
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Re: Isosceles Triangle
« Reply #8 on: May 15th, 2003, 4:59pm » |
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I used a picture to explain the possible points:
Let me know if I'm wrong.
-- Center of the circle is the intersection of the lines, n's.
[/u]
[u]Previous Answers:
_=-'-=_ All points on Ray A to Y (and beyond). (not including point A)
_=-'-=_ All points on Ray C to Z (and beyond). (not including point C)
_=-'-=_ All points on Ray B to X (and beyond). (not including point B)
_=-'-=_ All points on Arc AC (the lower arc). (not including points A and C)
By the way, the cirlce's radius only has to equal from the center to either point A or C ... B seems to be unimportant.
// image link edited by administrator to bypass remote file linking restriction on lightboxes.4t.com
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| « Last Edit: May 15th, 2003, 7:10pm by william wu » |
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Leo Broukhis
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Re: Isosceles Triangle
« Reply #9 on: May 15th, 2003, 5:20pm » |
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on May 15th, 2003, 4:59pm, Lightboxes wrote:I used a picture to explain the possible points:
Let me know if I'm wrong.
[u]Previous Answers:
_=-'-=_ All points on Ray B to X (and beyond). (not including point B)
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There is nothing wrong with points on line BX above B, they should be included as well.
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_=-'-=_ All points on Arc AC (the lower arc). (not including points A and C) |
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The arc part is right, if ABC is inscribed in the circle, as in the picture.
Quote:| By the way, the cirlce's radius only has to equal from the center to either point A or C ... B seems to be unimportant. |
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That I don't understand. B is important in finding the center of the circle, isn't it?
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Lightboxes
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Re: Isosceles Triangle
« Reply #10 on: May 15th, 2003, 8:45pm » |
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yeah, I was in a hurry so I could solve other riddles 
I missed the details...but with your comments and mine...I can't see any more answers on the plane.
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James Fingas
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Re: Isosceles Triangle
« Reply #11 on: May 16th, 2003, 12:26pm » |
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For any line emanating from B (except the one bisecting AC), I think we'll only get one solution. So I think those three answers are it. Here's my explanation of the arc:
Draw the circle which passes through A,B, and C. Now I suppose you know that when you fix two points (A and C) at opposite ends of a circle's diameter, and then pick any other point (M) on the circle, the angle AMC is always 90 degrees.
More generally, if you choose a circular arc with endpoints A and C, then the angle AMC is the same for all points M on the arc.
Now we know that for the circle passing through A,B, and C there is at least one point M for which AMB = CMB (the one on the line bisecting AC). This occurs on the arc between A and C that doesn't pass through point B. But on this arc, the angles AMB and AMC are constant (because it is part of the arc with endpoints AB that passes through point C, and part of the arc with endpoints BC that passes through A). Because AMB and AMC are constant, and we know AMB=AMC at one point on the arc, AMB=AMC at all points on the arc.
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Void_Myle
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Re: Isosceles Triangle
« Reply #12 on: May 17th, 2003, 10:02pm » |
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Do u people seriously have nothin better too do AT ALL!! like have you ever left the house have you heard of the sun!?
[font=Verdana][/MYLE[color=Red][/color]]
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ThudnBlunder
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Re: Isosceles Triangle
« Reply #13 on: May 18th, 2003, 1:27am » |
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Quote:| Do u people seriously have nothin better too do AT ALL!! like have you ever left the house have you heard of the sun!? |
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How did you get the name 'Void'?
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Void_Myle
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Re: Isosceles Triangle
« Reply #14 on: May 18th, 2003, 9:04am » |
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BEcuz i have no idea or nothin but i am pretty smart im just not very bright!   
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ThudnBlunder
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Re: Isosceles Triangle
« Reply #15 on: May 18th, 2003, 9:47am » |
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Quote:| ...but i am pretty smart... |
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I want to be as smart as you, so...............please tell me where you buy your clothes.
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| « Last Edit: May 18th, 2003, 12:40pm by ThudnBlunder » |
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Lightboxes
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Re: Isosceles Triangle
« Reply #16 on: May 18th, 2003, 8:58pm » |
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lol Thud...anyway...
Quote:Do u people seriously have nothin better too do AT ALL!! like have you ever left the house have you heard of the sun!?
[font=Verdana][/MYLE] |
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I was the one who did the picture...and it only took me 10-15 min. for the solution (autocad, I kinda cheated because I didn't do any math...I just moved the point around ALL OVER the plane until I found equal angles lol) and the picture took only 15-20 min. I don't see how solving this problem can take up my entire life.
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