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

Labels:
100 degrees,
30 degrees,
50,
area,
inradius,
measurement,
metric relations,
triangle

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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

https://goo.gl/photos/mJcfykipaSNAr2en6

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)