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Source: Indian Postal Coaching 2005

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1592 posts

Rushil

#1 h Sep 22, 2005, 3:42 PM • 3 Y

Y by Adventure10, Mango247, Mango247

Characterize all triangles $ABC$ s.t.
$AI_a : BI_b : CI_c = BC: CA : AB$ where $I_a$ etc. are the corresponding excentres to the vertices $A, B , C$

This post has been edited 1 time. Last edited by Rushil, Oct 15, 2005, 1:49 PM

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#2 h Sep 23, 2005, 2:28 PM • 2 Y

Y by Adventure10, Mango247

I think there is the following relation, isn't it??

$sin\frac{A}{2}=\frac{r_a}{AI_a}$ where $r_a$ is the radious of the excircle!!

Now if we substitute and use $r_a=p \cdot tg\frac{A}{2}$
I think we find that it is equilateral!

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Rushil

#3 h Sep 23, 2005, 2:29 PM • 2 Y

Y by Adventure10, Mango247

I believe we get some more cases-- not sure!!!

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#4 h Sep 23, 2005, 9:01 PM • 1 Y

Y by Adventure10

$\cos \frac A2=\frac {s}{AI_a}\Longleftrightarrow AI_a=\frac {s}{\cos \frac A2}$ . Thus, $\frac {AI_a}{a}=\frac {BI_b}{b}=\frac {CI_c}{c}\Longleftrightarrow a\cos \frac A2=b\cos \frac B2=c\cos \frac C2\Longleftrightarrow$
$a^3(s-a)=b^3(s-b)=c^3(s-c)\Longleftrightarrow a=b=c.$

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#5 h Nov 26, 2005, 9:47 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

There are indeed "some more cases", what is because $a^3\left(s-a\right)=b^3\left(s-b\right)=c^3\left(s-c\right)$ does not imply a = b = c.

Anyway, let me give a complete solution of the problem.

Theorem 1. Let ABC be a triangle, and let $I_a$ , $I_b$ and $I_c$ be the centers of its A-excircle, B-excircle and C-excircle, respectively. Then, we have

$AI_a: BI_b: CI_c=a: b: c$

if and only if

- either a = b = c,
- or b = c and $a=\frac{1+\sqrt5}{2}b$ ,
- or c = a and $b=\frac{1+\sqrt5}{2}c$ ,
- or a = b and $c=\frac{1+\sqrt5}{2}a$ .

Proof of Theorem 1. First, let's consider an arbitrary triangle ABC.

Let the A-excircle of triangle ABC touch the line CA at a point T. Then, since $I_a$ is the center of this A-excircle, $I_aT\perp CA$ , so the triangle $I_aTA$ is right-angled at T; also, $\measuredangle I_aAT=\frac{A}{2}$ (since the point $I_a$ , being the center of the A-excircle of triangle ABC, lies on the angle bisector of the angle CAB), and AT = s, where $s= \frac{a+b+c}{2}$ is the semiperimeter of triangle ABC. Thus, in the right-angled triangle $I_aTA$ , we have $AI_a=\frac{AT}{\cos\measuredangle I_aAT}$ , so that $AI_a=\frac{s}{\cos\frac{A}{2}}$ . But by the half-angle formulas, $\cos\frac{A}{2}=\sqrt{\frac{s\left(s-a\right)}{bc}}$ . Thus,

$AI_a=\frac{s}{\sqrt{\frac{s\left(s-a\right)}{bc}}}=\sqrt{\frac{sabc}{a\left(s-a\right)}}$ .

Similarly, $BI_b=\sqrt{\frac{sabc}{b\left(s-b\right)}}$ and $CI_c=\sqrt{\frac{sabc}{c\left(s-c\right)}}$ . Hence, we can equivalently transform the relation $AI_a: BI_b: CI_c=a: b: c$ as follows:

$AI_a: BI_b: CI_c=a: b: c$
$\Longleftrightarrow\ \ \ \ \ \sqrt{\frac{sabc}{a\left(s-a\right)}}: \sqrt{\frac{sabc}{b\left(s-b\right)}}: \sqrt{\frac{sabc}{c\left(s-c\right)}}=a: b: c$
$\Longleftrightarrow\ \ \ \ \ \sqrt{\frac{1}{a\left(s-a\right)}}: \sqrt{\frac{1}{b\left(s-b\right)}}: \sqrt{\frac{1}{c\left(s-c\right)}}=a: b: c$
$\Longleftrightarrow\ \ \ \ \ \frac{1}{a\left(s-a\right)}: \frac{1}{b\left(s-b\right)}: \frac{1}{c\left(s-c\right)}=a^2: b^2: c^2$
$\Longleftrightarrow\ \ \ \ \ 1: 1: 1=\left(a^3\left(s-a\right)\right): \left(b^3\left(s-b\right)\right): \left(c^3\left(s-c\right)\right)$
$\Longleftrightarrow\ \ \ \ \ a^3\left(s-a\right)=b^3\left(s-b\right)=c^3\left(s-c\right)$ .

Since $s-a=\frac{a+b+c}{2}-a=\frac{b+c-a}{2}$ and similarly $s-b=\frac{c+a-b}{2}$ and $s-c=\frac{a+b-c}{2}$ , this becomes

$a^3\left(b+c-a\right)=b^3\left(c+a-b\right)=c^3\left(a+b-c\right)$ .

Now, let ABC be a triangle satisfying the condition $AI_a: BI_b: CI_c=a: b: c$ . Then, we must therefore have $a^3\left(b+c-a\right)=b^3\left(c+a-b\right)=c^3\left(a+b-c\right)$ . Now, since our situation is symmetric, we can WLOG assume that $a\geq b\geq c$ . Since $b^3\left(c+a-b\right)=c^3\left(a+b-c\right)$ , we have

$0=b^3\left(c+a-b\right)-c^3\left(a+b-c\right)=a\left(b^3-c^3\right)-\left(b^4-c^4-b^3c+c^3b\right)$
$=a\left(b^2+bc+c^2\right)\left(b-c\right)-\left(b^3+c^3\right)\left(b-c\right)$
$=\left(a\left(b^2+bc+c^2\right)-\left(b^3+c^3\right)\right)\left(b-c\right)$ .

Now,

$a\left(b^2+bc+c^2\right)-\left(b^3+c^3\right)=\underbrace{\left(a-b\right)}_{\geq 0\text{, since }a\geq b}b^2+\underbrace{\left(a-c\right)}_{\geq 0\text{, since }a\geq c}c^2+\underbrace{abc}_{>0}>0$ ;

thus, we must have b - c = 0, so that b = c. Hence, the equation $a^3\left(b+c-a\right)=b^3\left(c+a-b\right)$ becomes $a^3\left(b+b-a\right)=b^3\left(b+a-b\right)$ , i. e. $a^3\left(2b-a\right)=b^3a$ . Division by a transforms this into $a^2\left(2b-a\right)=b^3$ , and thus

$0=a^2\left(2b-a\right)-b^3=\left(a-b\right)\left(ab-a^2+b^2\right)$ .

Hence, either a - b = 0, what leads to a = b and thus to a = b = c, or $ab-a^2+b^2=0$ , what is a quadratic equation in a and has the solutions $a=\frac{1+\sqrt5}{2}b$ and $a=\frac{1-\sqrt5}{2}b$ , of which the second one can be excluded since it is negative (and sidelengths of a triangle cannot be negative), so we get the solution b = c and $a=\frac{1+\sqrt5}{2}b$ . The other solutions can be obtained similarly (we WLOG assumed that $a\geq b\geq c$ and thus didn't get the cyclic permutations). By tracing the above argumentation backwards, we see that these solutions are indeed solutions.

Theorem 1 is proven.

Darij

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#6 h Nov 27, 2005, 3:04 PM • 1 Y

Y by Adventure10

Darij, you are true. I was negligently (it is regrettably !) I will prove that at least two between $a,b,c$ are equally, i.e. $(a-b)(b-c)(c-a)=0$ . Thus,
$a\ne b\ne c\ne a\ \wedge \ a^3(b+c-a)=b^3(c+a-b)=c^3(a+b-c)\Longrightarrow$
$a^3+b^3=c(a^2+ab+b^2),\ b^3+c^3=a(b^2+bc+c^2),\ c^3+a^3=b(c^2+ca+a^2)\Longrightarrow$
$a^2+b^2+c^2=0$ , what is absurd. Therefore, $a=b\ \vee\ b=c\ \vee\ c=a$ a.s.o.

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Rushil

#7 h Nov 28, 2005, 4:52 AM • 3 Y

Y by Adventure10, Adventure10, Mango247

Ya , darij! I got the same answers... Those were the other cases I was talking about above!

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