Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.
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Thursday, November 24, 2016
Geometry Problem 1288 Triangle, 30-50-100 Degrees, Area, Inradius, Metric Relations, Measurement
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Problem 1288
ReplyDeleteDraw BK perpendicular in AC,then BK=AB/2=c/2 and AK=√3c/2.
Let area triangle ABC is (ABC)=(AC.BK)/2=bc/4.
Let s=(a+b+c)/2 then (ABC)=sr.So r=(ABC)/s=bc/4s, suffices to show that bc/4s=c^2/2(b+c) or b/(a+b+c)=c/(b+c) or b^2=c(a+c).
In extension of the AB passes point D such that BD=BC=a,then triangle ABC is similar with triangle ACD so AB/AC=AC/AD or c/b=b/(a+c) or b^2=c(a+c).So r=(c^2)/(2(b+c)).
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE
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ReplyDeleteDraw CD such that angle BCD= 50 . D is on AB extended (see sketch).
1. S(ABC) =1/2.AC.AB.sin(30)= b.c/4
2. Note that angle BDC= 50 and tri CBD is isosceles => BD=BC= a
Triangle ABC and ACD are similar => AC/AD=AB/AC= c/b => AD= b^2/c
BD= AD-AB= b^2/c-c= a
AB/AC=BC/CD=c/b => CD=a.b/c= b(b^2/c^2-1)
Inradius r= S(ABC)/2p where 2p= a+b+c= b+b^2/c
. r=b.c/4p= bc/2/(b+b^2/c) => r=c^2/(2(b+c))
3. Apply cosine formula to triangle ACD
CD^2=AC^2+AD^2-2.AC.AD.cos(30)…..(1)
Replace CD=b(b^2/c^2-1) ; AD= b^2/c in (1) and simplifying we have
. (b^2/c^2-1)^2=1+b^2/c^2-(b/c).sqrt(3)
This equation will simplify to b^3+sqrt(3).c^3= 3.b.c^2
Draw a perpendicular from B to AC and meet it at D
ReplyDeleteConsider the 30-60-90 triangle ABD
Since AB = c, BD = c/2, AD = Sqrt(3)c/2
Hence Area of ABC = bc/4 ---------- (1)
Draw an Angle Bisector for m(AB,BC) and let it meet AC at E
From Angle bisector theorem, CE = ab/a+c and AE = bc/a+c
Since triangle BEC is isosceles (50,50,80), BE = CE = ab/a+c ---------(2)
If you observe, triangle ABC is similar to AEB
Hence BE/c=a/b => BE = ac/b --------------(3)
From (2) and (3) ab/a+c = ac/b
hence a = b^2-c^2/c -------(4)
As we know semi-perimeter for triangle ABC, S = a+b+c/2
substituting value of 'a' from (4), we get S = b^2+bc/2c
We know, Area ABC in terms of S and r = S.r = bc/4 (from (1))
=> (b^2+bc/2c ).r = bc/4
=> r= c^2/2(b+c) ----------- (5)
Now consider the right triangle BED
We know BD = c/2 , BE = b^2-c^2/2b (Substitute for 'a' in ac/b)
and ED = AD - AE = Sqrt(3)bc-2c^2/2b
Applying pythogorus, we get BE^2 = BD^2+ED^2
=> (b^2-c^2/b)^2 = (Sqrt(3)bc-2c^2/2b)^2 + c^2/4
Simplifying, we get b^3+Sqrt(3)c^3=3b^2c^2 ----------------(6)