wu :: forums « wu :: forums - Integral Square » Welcome, Guest. Please Login or Register. Apr 30th, 2024, 1:24pm

RIDDLES SITE WRITE MATH! Home Help Search Members Login Register

wu :: forums    riddles    easy (Moderators: ThudnBlunder, Eigenray, SMQ, towr, Icarus, william wu, Grimbal)    Integral Square « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: Integral Square  (Read 1109 times) Sir Col Uberpuzzler impudens simia et macrologus profundus fabulae     Gender: Posts: 1825 Integral Square   « on: Aug 29th, 2007, 9:55am » Quote Modify A right triangle has legs with integer lengths x and y. A square is placed inside the triangle with one vertex at the right angle and the opposite vertex on the hypotenuse. Determine the conditions for the square to have an integral side length. IP Logged mathschallenge.net / projecteuler.net ThudnBlunder wu::riddles Moderator Uberpuzzler The dewdrop slides into the shining Sea     Gender: Posts: 4489 Re: Integral Square   « Reply #1 on: Aug 29th, 2007, 10:08am » Quote Modify x+y divides xy IP Logged THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you. ecoist Senior Riddler     Gender: Posts: 405 Re: Integral Square   « Reply #2 on: Aug 31st, 2007, 3:24pm » Quote Modify The existence alone of a square so located inside a right triangle restricts x and y. IP Logged Sameer Uberpuzzler Pie = pi * e     Gender: Posts: 1261 Re: Integral Square   « Reply #3 on: Aug 31st, 2007, 4:56pm » Quote Modify I get If a is length of square, then a2 = (x-a)*(y-a) . I can't think of what to express it  in words yet... IP Logged "Obvious" is the most dangerous word in mathematics. --Bell, Eric Temple Proof is an idol before which the mathematician tortures himself. Sir Arthur Eddington, quoted in Bridges to Infinity ecoist Senior Riddler     Gender: Posts: 405 Re: Integral Square   « Reply #4 on: Sep 1st, 2007, 6:27pm » Quote Modify What integers satisfy ThundanBlunder's condition?  If a and b are arbitrary positive integers, then x=a(a+b) and y=b(a+b) satisfies his condition, with a side of the inscribed square of length ab.  But, alas, x=18 and y=9 also satisfies ThudanBlunder's condition, with square of side length 6, yet the above formula doesn't cover this case. IP Logged Eigenray wu::riddles Moderator Uberpuzzler     Gender: Posts: 1948 Re: Integral Square   « Reply #5 on: Sep 1st, 2007, 7:13pm » Quote Modify An integer parametrization is x = ka(a+b), y = kb(a+b), where gcd(a,b)=1. IP Logged Sameer Uberpuzzler Pie = pi * e     Gender: Posts: 1261 Re: Integral Square   « Reply #6 on: Sep 1st, 2007, 7:21pm » Quote Modify on Sep 1st, 2007, 6:27pm, ecoist wrote: yet the above formula doesn't cover this case.   I see. It's weird. I started with this: Square of length a x = a+b y = a+c So sqrt(a2 + c2) + sqrt(a2 + b2) = sqrt( (a+c)2 + (a+b)2)   Simplifying which gives me a2 = bc   I wonder why doesn't the case of x=18 and y = 9  get covered!!   « Last Edit: Sep 1st, 2007, 7:22pm by Sameer » IP Logged "Obvious" is the most dangerous word in mathematics. --Bell, Eric Temple Proof is an idol before which the mathematician tortures himself. Sir Arthur Eddington, quoted in Bridges to Infinity Grimbal wu::riddles Moderator Uberpuzzler     Gender: Posts: 7527 Re: Integral Square   « Reply #7 on: Sep 2nd, 2007, 2:53pm » Quote Modify on Sep 1st, 2007, 7:21pm, Sameer wrote: I wonder why doesn't the case of x=18 and y = 9  get covered!!   Doesn't it? a = 6, b = 3, c = 12 a2 = b·c = 36 x = a+b = 9, y = a+c = 18 « Last Edit: Sep 2nd, 2007, 3:15pm by Grimbal » IP Logged Sameer Uberpuzzler Pie = pi * e     Gender: Posts: 1261 Re: Integral Square   « Reply #8 on: Sep 2nd, 2007, 3:00pm » Quote Modify on Sep 2nd, 2007, 2:53pm, Grimbal wrote: Isn't it? a = 6, b = 3, c = 12 a2 = b·c = 36 x = a+b = 9, y = a+c = 18   Duh!?! Thanks!! Seems like I have been overworking again!! However my original question still holds. How do you geometrically or in words express a2 = bc!! IP Logged "Obvious" is the most dangerous word in mathematics. --Bell, Eric Temple Proof is an idol before which the mathematician tortures himself. Sir Arthur Eddington, quoted in Bridges to Infinity Grimbal wu::riddles Moderator Uberpuzzler     Gender: Posts: 7527 Re: Integral Square   « Reply #9 on: Sep 2nd, 2007, 3:01pm » Quote Modify Inserting the square creates 2 smaller triangles, one with sides b and a, the other with sides a and c.  The fact that they are similar implies b/a = a/c, which directly gives a2 = bc « Last Edit: Sep 2nd, 2007, 3:02pm by Grimbal » IP Logged Sir Col Uberpuzzler impudens simia et macrologus profundus fabulae     Gender: Posts: 1825 Re: Integral Square   « Reply #10 on: Sep 2nd, 2007, 3:07pm » Quote Modify And using similarity, (x-a)/a = a/(y-b), which gives a2 = (x-a)(y-a) = xy-a(x+y)+a2, leading to T&B's result: a = xy/(x+y). IP Logged mathschallenge.net / projecteuler.net Grimbal wu::riddles Moderator Uberpuzzler     Gender: Posts: 7527 Re: Integral Square   « Reply #11 on: Sep 2nd, 2007, 3:09pm » Quote Modify The fact that the 2 smaller trianges are similar implies    (x-a)/a = a/(y-a)    <=> (x-a)(y-a) = a2    <=> xy-ax-ay-a2 = a2    <=> xy-a(x+y) = 0    <=> a = xy/(x+y) So, for a to be an integer, xy must be a multiple of (x+y), which conveniently removes the unknown a from the equation. And it proves what ThudanBlunder said straight away and Sir Col minutes ago.   « Last Edit: Sep 2nd, 2007, 3:12pm by Grimbal » IP Logged Sameer Uberpuzzler Pie = pi * e     Gender: Posts: 1261 Re: Integral Square   « Reply #12 on: Sep 2nd, 2007, 3:11pm » Quote Modify Thanks for the great simple explanation!!  I unnecessarily seem to make things complicated, don't I? IP Logged "Obvious" is the most dangerous word in mathematics. --Bell, Eric Temple Proof is an idol before which the mathematician tortures himself. Sir Arthur Eddington, quoted in Bridges to Infinity Sir Col Uberpuzzler impudens simia et macrologus profundus fabulae     Gender: Posts: 1825 Re: Integral Square   « Reply #13 on: Sep 2nd, 2007, 3:12pm » Quote Modify on Sep 2nd, 2007, 3:09pm, Grimbal wrote: (sigh!) At least you proved it. I think that T&B's was just a lucky guess.   IP Logged mathschallenge.net / projecteuler.net ThudnBlunder wu::riddles Moderator Uberpuzzler The dewdrop slides into the shining Sea     Gender: Posts: 4489 Re: Integral Square   « Reply #14 on: Sep 3rd, 2007, 7:59pm » Quote Modify on Sep 2nd, 2007, 3:12pm, Sir Col wrote: At least you proved it. I think that T&B's was just a lucky guess.   And the great thing is the more I study, the luckier I get.   IP Logged THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you. Pages: 1  Reply Notify of replies Send Topic Print Forum Jump: ----------------------------- riddles -----------------------------=> easy   - medium   - hard   - what am i   - what happened   - microsoft   - cs   - putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »

Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright © 2000-2004 Yet another Bulletin Board