Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Sunday, March 20, 2016
Geometry Problem 1201: Triangle, Interior Perpendicular Bisector, Midpoint, 90 Degrees, Area
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angle(BCA)=θ
ReplyDeletetriangle ABC ・・・cosθ=(a^2+b^2-c^2)/(2*a*b) ①
triangle DMC ・・・cosθ=(b/2)/DC=b/(2*DC) ②
①=②
DC=a*b^2/(a^2+b^2-c^2) ③
area triangle DMC=S*(1/2)*(DC/a)=(b/2)*DM*(1/2)
∴DM=2*S*DC/(a*b)
=2*S*③/(a*b)=a*b^2/(a^2+b^2-c^2)
S = bh/2...(1) where h is the length of the altitude from B
ReplyDeleteFrom similar triangles
d/h = b/2/sqrt(a^2-h^2)...(2) where d = MD
Eliminating h between (1) and (2),
d = S/sqrt(a^2-4S^2/b^2) = 2bs / sqrt{4a^2b^2 - 16s(s-a)(s-b)s-c)} where s =(a+b+c)/2 which when simplifying the algebra using the difference of 2 squares yields
d = MD = 2bS/(a^2+b^2-c^2)
Sumith Peiris
Moratuwa
Sri Lanka
Draw altitude BH
ReplyDeletePer law of cosine in triangle ABC we have Cos(C)= (a^2+b^2-c^2)/(2.a.b)
So CH= a. cos(C )= (a^2+b^2-c^2)/(2.b)
Triangles CBH similar to CDM so DM/BH= DM/DH= b^2/(a^2+b^2-c^2)
Replace BH= 2.S/b in above expression we get DM= 2.b.S/(a^2+b^2-c^2)
A simpler proof
ReplyDeleteLet H be the foot of the perpendicular from B to AC and let CH = e
Then applying Pythagoras to triangles ABH and CBH
c^2 - (b-e)^2 = a^2 - e^2 from which
e = ( a^2+b^2-c^2)/2b
From similar triangles
e/(b/2) = h/d where d = MD and h = BH
So (a^2+b^2-c^2)/b^2 = S/(b/2)/d
Hence MD = d = 2bS / (a^2+b^2-c^2)
Sumith Peiris
Moratuwa
Sri Lanka
S=1/2absinC=1/2abMD/DC=MD.a.MC/DC=MD.a.cosC=MD.a.(a^2+b^2-c^2)/2ab.
ReplyDeleteMD = MC tan C = (b/2) tan C = (b sin C)/(2 cos C) = (1/2) ab sin C)/(a cos C)
ReplyDelete= S/a cos C) = 2bS/2ab cos C = 2bS/(a^2+b^2-c^2)
RHS=2abS/a(a^2+b^2-c^2)=S/a.cosC=ab.sinC/2a.cosC= tanC/(b/2)=tanC/MC=MD
ReplyDelete