heights and inner radius of a triangle (BWM1988_1_2)

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heights and inner radius of a triangle (BWM1988_1_2)

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Source: Bundeswettbewerb Mathematik 1988, first stage, second question

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orl

#1 h Sep 6, 2003, 1:16 AM • 3 Y

Y by Adventure10, Mango247, EntropiaAwake

Let $h_a$ , $h_b$ and $h_c$ be the heights and $r$ the inradius of a triangle.
Prove that the triangle is equilateral if and only if $h_a + h_b + h_c = 9r$ .

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Soarer

#2 h Sep 6, 2003, 5:10 AM • 2 Y

Y by Adventure10, Mango247

aha=bhb=chc=2S=2rp
ha=2S/a
hb=2S/b
hc=2S/c
r=S/p
ha+hb+hc=2S/a+2S/b+2S/c=9r=9S/p
(1/a+1/b+1/c)=9/(a+b+c)
(a+b+c)(1/a+1/b+1/c)=9
By cauchy, (a+b+c)(1/a+1/b+1/c)>=9
So a/(1/a)=b/(1/b)=c/(1/c) or a=b=c

If a=b=c,
S=sqrt3 /4 a^2
ha=hb=hc=sqrt3/2 a
(3a/2)r=sqrt3/4 a^2
r=1/2sqrt3 a^2

ha+hb+hc=3sqrt3/2 a
9r=9/2sqrt3 a^2
=3sqrt3/2 a^2
So they are the same

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Eckard

#3 h Sep 6, 2003, 11:07 AM • 2 Y

Y by Adventure10, Mango247

Only a short addition: This was nearly the same problem as in the German Mathematical Olympiad 1998/99, grade 10, round 3.

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orl

#4 h Sep 6, 2003, 6:32 PM • 2 Y

Y by Adventure10, Mango247

Yeah, you're right. It was used in the Bundeswettbewerb Mathematik 1988 as second question in the first stage, for short BWM1988_1_2

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#5 h Jul 21, 2023, 7:15 PM

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ha+hb+HC=9r
we know, rs=∆
∆=(a/2)*ha
ha=(2∆)/a
Similarly,
hb=(2∆)/b
hc=(2∆)/c
Add,
2∆{(1/a)+(1/b)+(1/c)}=9r
2rs{(1/a)+(1/b)+(1/c)}=9r
2s{(1/a)+(1/b)+(1/c)}=9
(a+b+c){(1/a)+(1/b)+(1/c)}=9
So, after simply,
a²c+c²a+a²b+b²a+b²c+c²b=6abc
So,
(a√c-b√c)²+(c√a-b√a)²+(a√b-c√b)²=0
Sum of square of 3 numbers is 0 , so each must be equal to 0
So,
(a√c-b√c=0, a=b
c√a-b√a=0, c=b
a√b-c√b=0, a=c
so, a=b=c
Therefore triangle is equilateral.
#Krishijivi

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