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December 26th, 2012, 01:06 PM   #1
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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
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December 27th, 2012, 09:13 AM   #2
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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
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December 27th, 2012, 10:58 AM   #3
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Re: find area of BDC

you really love integers don't you
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December 29th, 2012, 12:39 PM   #4
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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...
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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 can be proven
but i'll ask that in a separate post
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December 30th, 2012, 09:51 AM   #6
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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
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December 31st, 2012, 07:36 AM   #7
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Re: find area of BDC

yes but just how you found those 3 equations, how you got to them?
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December 31st, 2012, 09:13 AM   #8
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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
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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
File Type: jpg Area of BCD.jpg (63.3 KB, 195 views)
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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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