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Topic: Concurrence in the Bride's Chair (Read 632 times) |
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ecoist
Senior Riddler
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Concurrence in the Bride's Chair
« on: Sep 5th, 2006, 8:12am » |
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The classic figure used to prove Pythagorus' Theorem is a right triangle ABC, with right angle at B and squares erected externally on its sides. The square on side AB is AA'B'B, that on side BC is BB''C''C, and that on side CA is CC'''A'''A (all vertices read clockwise). Additional lines in the figure are A'C, AC'', and the perpendicular to AC through B.
It turns out that these three additional lines are concurrent. Why?
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Deedlit
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Re: Concurrence in the Bride's Chair
« Reply #1 on: Sep 6th, 2006, 2:41pm » |
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Classical geometry problems were always my bane, ever since high school contests.  So my natural inclination is a cheap proof like this:
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Let D be the base of the perpendicular from B to AC. Place a coordinate axis with the origin at D, and with C on the positive x-axis. Choose x and c so that A = (-x, 0), B = (0, cx), and C = (xc^2, 0). Then A' = (-x - cx, x), and C'' = (cx + xc^2, xc^2). Calculating the y-intercepts for AC'' and A'C yields xc^2/(1 + c + c^2) for both of them.
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ecoist
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Re: Concurrence in the Bride's Chair
« Reply #2 on: Sep 6th, 2006, 9:47pm » |
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Very good, Deedlit! I didn't expect your solution. Would you believe I used Desargue's Theorem to solve it?
By the way, Ceva's Theorem has a "cheap" solution too. So maybe your solution is not that cheap.
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Barukh
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Re: Concurrence in the Bride's Chair
« Reply #3 on: Sep 7th, 2006, 3:36am » |
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Wow! I worked hard on solving this and completely forgot about Ceva’s theorem which gives an immediate answer.
It is interesting to see the proof that uses Desargue’s theorem. Theorems from projective geometry seem to me most natural for problems about concurrency.
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ecoist
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Re: Concurrence in the Bride's Chair
« Reply #4 on: Sep 7th, 2006, 9:41am » |
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Double wow! Ceva's Theorem solves this problem too? Here's my proof using Desargues' Theorem.
| hidden: | | The perpendicular to AC from B meets the intersection E of the lines A'B' and B''C'' because triangles ABC and BB'E are congruent. Consider the two triangles A'EC'' and ABC. Sides AB and EC'' meet at B''. Sides BC and A'E meet at B', Sides AC and A'C'' at, say, F. The three points, B', C'', and F are collinear since the triangles A'B'F and A'AF (or B''C''F and CC''F, depending on where F is) are congruent. Hence triangles A'EC'' and ABC are perspective with the line B'F, and so are also perspective with a point, namely, that point which is the intersection of the lines A'C, AC'', and ED. |
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| « Last Edit: Sep 7th, 2006, 2:20pm by ecoist » |
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Barukh
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Re: Concurrence in the Bride's Chair
« Reply #5 on: Sep 8th, 2006, 12:25am » |
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on Sep 7th, 2006, 9:41am, ecoist wrote:| Here's my proof using Desargues' Theorem. |
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Very nice! 
Just a little clarification: "The three points, B', C'', and F are collinear " should be The three points, B', B'', and F are collinear".
Quote:| Double wow! Ceva's Theorem solves this problem too? |
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| hidden: | | Let D be the projection of B on AC; P the intersection of AB with A’C; Q the intersection of BC with AC”. Now we consider the triangle ABC, where we denote BC = a, AB = c. Using similar triangles, we easily obtain that BP:AB = a/(a+c), BQ:BC = c/(a+c). Also, considering the areas of similar triangles ABD and BDC with a common leg BD, we get AD : DC = c2/a2. Now, Ceva’s theorem may be applied. |
This should not be a surprise: one of the classical proofs of Desargues’s theorem uses Ceva’s theorem.
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| « Last Edit: Sep 8th, 2006, 12:28am by Barukh » |
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ecoist
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Re: Concurrence in the Bride's Chair
« Reply #6 on: Sep 8th, 2006, 9:42am » |
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Thanks, Barukh, and thanks further for correcting my error.
Never knew Ceva's Theorem is if and only if. Nice concise proof.
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