wu :: forums « wu :: forums - Smallest Equilateral Triangle » Welcome, Guest. Please Login or Register. Nov 28th, 2024, 12:47pm

RIDDLES SITE WRITE MATH! Home Help Search Members Login Register

wu :: forums    riddles    medium (Moderators: SMQ, Eigenray, Icarus, towr, Grimbal, william wu, ThudnBlunder)    Smallest Equilateral Triangle « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: Smallest Equilateral Triangle  (Read 573 times) ThudnBlunder wu::riddles Moderator Uberpuzzler The dewdrop slides into the shining Sea     Gender: Posts: 4489 Smallest Equilateral Triangle   « on: Nov 28th, 2009, 2:51am » Quote Modify Find the area of the smallest equilateral triangle inscribable in a right triangle of legs a and b. « Last Edit: Dec 3rd, 2009, 5:54am by ThudnBlunder » IP Logged THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you. Immanuel_Bonfils Junior Member     Posts: 114 Re: Smallest Equilateral Triangle   « Reply #1 on: Nov 30th, 2009, 6:39pm » Quote Modify ABC is the right triangle and PPPthe inscribed equilateral with vetices respecively in AB, BC and CA, side length x. Let ABC = (constant or known: sin =b/(a+ b) ) and PPC=(variable or unknown). Sinus law in triangle PBPleads to    x = a.sin /(sin(/3+-)+cos.sin)  (*),   that has a minimum for cot =(sin+sin(/3-))/cos(/3-)).   Substitution  in (*) gives us xthat multiplied by (3)/4 leads to the required area. IP Logged ThudnBlunder wu::riddles Moderator Uberpuzzler The dewdrop slides into the shining Sea     Gender: Posts: 4489 Re: Smallest Equilateral Triangle   « Reply #2 on: Dec 1st, 2009, 2:49am » Quote Modify I will bear that in mind, but I was looking for a simple formula f(a,b). IP Logged THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you. SMQ wu::riddles Moderator Uberpuzzler     Gender: Posts: 2084 Re: Smallest Equilateral Triangle   « Reply #3 on: Dec 1st, 2009, 9:02am » Quote Modify Place the right angle at the origin with leg a along the positive y-axis and leg b along the positive x-axis.  Pick a point X = (x, 0) on side b.  If we pick another point P anywhere on the triangle, there are two points Q and Q' which will form an equilateral triangle with X and P.  The sets of Q and Q' over all P form images of the triangle rotated by +/-120o around point X.  To find the inscribed equilateral triangle containing X we can find point Y = (0, y): the intersection of the image of the hypotenuse (rotated +120o) with leg a, i.e. the y-axis.  With some algebra, we obtain y = [(b - x)3]/2 - (b + x)(b3 - a)/[2(b + a3)].   We now seek to minimize XY = (x2 + y2).  Equivalently, and more simply, we can minimize XY2 = x2 + y2.  Setting d/dx (x2 + y2) = 0 gives, following some more algebra, x2 + y2 = a2b2/(a2 + ab3 + b2).   Finally, the area of the inscribed equilateral triangle = (XY23)/4 = [(x2 + y2)3]/4, so the area of the minimal inscribed equilateral triangle is (a2b23)/[4(a2 + ab3 + b2)].   --SMQ IP Logged --SMQ ThudnBlunder wu::riddles Moderator Uberpuzzler The dewdrop slides into the shining Sea     Gender: Posts: 4489 Re: Smallest Equilateral Triangle   « Reply #4 on: Dec 1st, 2009, 10:51am » Quote Modify Excellent, SMQ, as always.   I will shortly post a solution that uses a neat little substitution to simplify matters. Perhaps even a diagram, but I am presently using Vista and most of my drawing software doesn't seem to want to work with it. Although Cabri doesn't seem very fussy. « Last Edit: Dec 1st, 2009, 10:53am by ThudnBlunder » IP Logged THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you. ThudnBlunder wu::riddles Moderator Uberpuzzler The dewdrop slides into the shining Sea     Gender: Posts: 4489 Re: Smallest Equilateral Triangle   « Reply #5 on: Dec 26th, 2009, 2:24pm » Quote Modify Sorry, no diagram.   But with a pencil and paper draw the right-angled triangle ABC, with AC as the hypotenuse. Let AB = a and BC = b Now draw the inscribed equilateral triangle DEF with D on AC, E on AB, and F on BC Let BF = b1 and FC = b2   Let DE = EF = FD = s Let angle EFB = Let angle DCF =   Then angle FDC = 60 + - and b = b1 + b2      = s*cos + [s*sin(60 + - )/sin]   Now letting  r =  3  + b/a and p = 3b/a + 1 we obtain 1/s = (rsin + pcos)/2b   We seek 1/s to be a maximum when we differentiate wrt to and setting d(1/s)/dto zero we obtain   tan = r/p and minimum value of s2 = a2b2/(a2 + 3ab + b2)     And required area = 3s2/4 =  (3/4)a2b2/(a2 + 3ab + b2)     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