My Math Forum find area of BDC

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 December 26th, 2012, 01:06 PM #1 Senior Member   Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11 find area of BDC 1) could someone find out the are of triangle BDC: given that A is the circumcenter point of BDC E is the incenter point of BAC and |EI| = 0.72 F is the incenter point of BAD and |FH| = 1 G is the incenter point of DAC and |GJ| = 0.6 NOTE: the angles D,B,C are smaller then 90° 2) find area of EFG ---------------------------------------------------------- answers: 1) 14.83 2) 1.1
 December 27th, 2012, 09:13 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: find area of BDC Sure wish you'd have triangle BCD on x-axis, with D at origin! Haven't done the solving yet (not sure if I will!) but here's one with all-integers: areaBCD = 290304 BD = 624 BC = 960 CD = 1008 AB = AC = AD = 520 FH = 156 (AF = 260) EI = 96 (AE = 104) GJ = 63 (AG = 65) In case you wanted to "play" with an integer case
 December 27th, 2012, 10:58 AM #3 Senior Member   Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11 Re: find area of BDC you really love integers don't you
 December 29th, 2012, 12:39 PM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: find area of BDC Code:  C M B A OK Jelly, using mine to "do the work"; problem statement: M is the circumcenter of acute triangle ABC, so AM = BM = CM. The inradius of triangle ABM = 63 The inradius of triangle ACM = 96 The inradius of triangle BCM = 156 Calculate sides of triangle ABC (same as yours, right?) I think this'll work: Apply rs = Area = sqrt(s(s-a)(s-b)(s-c)) to triangles ABM, ACM, BCM. Let AM = BM = CM = R. AB = 2 R sin(C) , BC = 2 R sin(A), AC = 2 R sin(B) R sin(C)cos(C) / (1 + sin(C)) = 63 R sin(B)cos(B) / (1 + sin (B)) = 96 R sin(A)cos(A) / (1 + sin(A)) = 156 A + B + C = pi. 4 equations, 4 unknowns: should be ok, right? Try it...
December 30th, 2012, 08:12 AM   #5
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Re: find area of BDC

Quote:
 Originally Posted by Denis R sin(C)cos(C) / (1 + sin(C)) = 63 R sin(B)cos(B) / (1 + sin (B)) = 96 R sin(A)cos(A) / (1 + sin(A)) = 156
i don't see how you get to those
and i don't see how $\frac {a}{sin(A)}= \frac {b}{sin(B)} = \frac {c}{sin(C)} = 2R$ can be proven
but i'll ask that in a separate post

 December 30th, 2012, 09:51 AM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: find area of BDC R sin(C)cos(C) / (1 + sin(C)) = 63 R sin(B)cos(B) / (1 + sin (B)) = 96 R sin(A)cos(A) / (1 + sin(A)) = 156 .................................................. .......................................... Angles are: A = 36.869897.... , B = 67.380135.... , C = 75.749967.... And R = 520. 520 * sin(75.749967) * cos(75.749967) / [1 + sin(75.749967)] = 63 520 * sin(67.380135) * cos(67.380135) / [1 + sin(67.380135)] = 96 520 * sin(36.869897) * cos(36.869897) / [1 + sin(36.869897)] = 156
 December 31st, 2012, 07:36 AM #7 Senior Member   Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11 Re: find area of BDC yes but just how you found those 3 equations, how you got to them? $R sin(C)cos(C) / (1 + sin(C)) = 63 R sin(B)cos(B) / (1 + sin (B)) = 96 R sin(A)cos(A) / (1 + sin(A)) = 156$
 December 31st, 2012, 09:13 AM #8 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: find area of BDC Forgot exactly where/how, but I use them as a "fait accompli"! I think you'll see why here: http://mathworld.wolfram.com/Inradius.html Bonne et heureuse annee
January 2nd, 2013, 03:28 AM   #9
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Re: find area of BDC

[attachment=0:2n7qvt2p]Area of BCD.jpg[/attachment:2n7qvt2p]
the calculation is too tedious
hope gelatine1 can post your solution
Attached Images
 Area of BCD.jpg (63.3 KB, 195 views)

January 2nd, 2013, 08:55 AM   #10
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Re: find area of BDC

Quote:
 Originally Posted by Denis Bonne et heureuse annee
merci et je vous souhaitez aussi une bonne année

Quote:
 Originally Posted by albert.teng hope gelatine1 can post your solution
i have no solution i just randomly made a question and found the answer with some program. just because it's an intresting problem and i am learning quite alot from this

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