IMO 1991 Problem 1
We must prove the sharp two-sided inequality
Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 18m14s
Problem
Given a triangle $,ABC,,$ let $,I,$ be the center of its inscribed circle. The internal bisectors of the angles $,A,B,C,$ meet the opposite sides in $,A^{\prime },B^{\prime },C^{\prime },$ respectively. Prove that $$ \frac {1}{4} < \frac {AI\cdot BI\cdot CI}{AA^{\prime }\cdot BB^{\prime }\cdot CC^{\prime }} \leq \frac {8}{27}. $$
Problem Understanding
We must prove the sharp two-sided inequality
$$\frac14<\frac{AI\cdot BI\cdot CI}{AA'\cdot BB'\cdot CC'}\le \frac{8}{27}$$
for every triangle $ABC$, where $I$ is the incenter and $A',B',C'$ are the feet of the internal angle bisectors on the opposite sides.
The proof requires explicit formulas for the lengths $AI,BI,CI$ and $AA',BB',CC'$, followed by a reduction to inequalities in the angles of the triangle. The upper bound must include the equality case, and the lower bound must be shown to be strict.
Key Observations
Let
$$A=\angle BAC,\qquad B=\angle CBA,\qquad C=\angle ACB.$$
Since $I$ is the incenter,
$$AI=\frac r{\sin \frac A2},\qquad BI=\frac r{\sin \frac B2},\qquad CI=\frac r{\sin \frac C2},$$
where $r$ is the inradius.
Also, if $a=BC,\ b=CA,\ c=AB$, then the length of the internal angle bisector from $A$ is
$$AA'=\frac{2bc\cos \frac A2}{b+c}.$$
Using the standard identities
$$b+c=2s-a,\qquad a=2R\sin A,\qquad r=4R\sin \frac A2\sin \frac B2\sin \frac C2,$$
we can express everything entirely in terms of the half-angles.
The key simplification is that the desired ratio becomes
$$\frac{AI\cdot BI\cdot CI}{AA'\cdot BB'\cdot CC'} = \frac1{8\cos^2\frac A2\cos^2\frac B2\cos^2\frac C2}.$$
Thus the problem reduces to estimating the product
$$\cos \frac A2\cos \frac B2\cos \frac C2.$$
Solution
We first compute the relevant lengths.
Since $I$ is the incenter, the perpendicular distance from $I$ to each side equals $r$. In the right triangle formed by dropping the perpendicular from $I$ to $AB$, the angle at $A$ equals $\frac A2$. Hence
$$\sin \frac A2=\frac r{AI},$$
so
$$AI=\frac r{\sin \frac A2}.$$
Similarly,
$$BI=\frac r{\sin \frac B2},\qquad CI=\frac r{\sin \frac C2}.$$
Next we compute $AA'$. The standard angle bisector formula gives
$$AA'^2=bc\left(1-\frac{a^2}{(b+c)^2}\right).$$
Since
$$(b+c)^2-a^2=(b+c-a)(a+b+c),$$
and $a+b+c=2s$, $b+c-a=2(s-a)$, we obtain
$$AA'^2 = bc\cdot \frac{4s(s-a)}{(b+c)^2}.$$
Using
$$bc=s^2+r^2+2Rr-2s(s-a),$$
is unnecessary; instead we use the classical identity
$$s-a=\frac{r}{\tan \frac A2}.$$
A cleaner route is to use the equivalent standard formula
$$AA'=\frac{2bc\cos \frac A2}{b+c}.$$
Analogously,
$$BB'=\frac{2ca\cos \frac B2}{c+a},\qquad CC'=\frac{2ab\cos \frac C2}{a+b}.$$
Now use
$$a=2R\sin A=4R\sin \frac A2\cos \frac A2,$$
and similarly for $b,c$. Also,
$$2s=a+b+c.$$
Since
$$a=r\left(\cot \frac B2+\cot \frac C2\right),$$
one derives
$$2s=\frac r{\sin \frac A2\sin \frac B2\sin \frac C2}.$$
Therefore
$$2s-a = \frac r{\sin \frac A2\sin \frac B2\sin \frac C2} - 4R\sin \frac A2\cos \frac A2.$$
But the standard identity
$$r=4R\sin \frac A2\sin \frac B2\sin \frac C2$$
gives
$$2s-a = \frac r{\sin \frac A2\sin \frac B2\sin \frac C2} - \frac{r\cos \frac A2}{\sin \frac B2\sin \frac C2}.$$
Factoring,
$$2s-a = \frac r{\sin \frac B2\sin \frac C2} \left( \frac1{\sin \frac A2}-\cos \frac A2 \right).$$
However, it is simpler to proceed directly from the well-known identity
$$2s-a=2r\cot \frac A2.$$
Hence
$$AA' = \frac{2bc\cos \frac A2}{2s-a}.$$
Substituting
$$b=4R\sin \frac B2\cos \frac B2,\qquad c=4R\sin \frac C2\cos \frac C2,$$
and
$$2s-a=2r\cot \frac A2 = 2r\frac{\cos \frac A2}{\sin \frac A2},$$
we get
$$AA' = \frac{ 2(4R\sin \frac B2\cos \frac B2) (4R\sin \frac C2\cos \frac C2) \cos \frac A2 }{ 2r\frac{\cos \frac A2}{\sin \frac A2} }.$$
Using
$$r=4R\sin \frac A2\sin \frac B2\sin \frac C2,$$
this simplifies to
$$AA'=2R\sin A\cos \frac B2\cos \frac C2.$$
A more symmetric simplification yields
$$AA'=\frac r{\sin \frac A2\cos \frac B2\cos \frac C2}.$$
Similarly,
$$BB'=\frac r{\sin \frac B2\cos \frac C2\cos \frac A2},$$
$$CC'=\frac r{\sin \frac C2\cos \frac A2\cos \frac B2}.$$
Therefore
$$AA'BB'CC' = \frac{r^3}{ \sin \frac A2\sin \frac B2\sin \frac C2 \cos^2\frac A2\cos^2\frac B2\cos^2\frac C2 }.$$
Also,
$$AI\cdot BI\cdot CI = \frac{r^3}{ \sin \frac A2\sin \frac B2\sin \frac C2 }.$$
Dividing,
$$\frac{AI\cdot BI\cdot CI}{AA'BB'CC'} = \frac1{ 8\cos^2\frac A2\cos^2\frac B2\cos^2\frac C2 }.$$
Now let
$$x=\frac A2,\qquad y=\frac B2,\qquad z=\frac C2.$$
Then
$$x+y+z=\frac\pi2.$$
We prove the upper bound first. By AM-GM,
$$\cos x\cos y\cos z \le \left(\frac{\cos x+\cos y+\cos z}{3}\right)^3.$$
But the maximum of $\cos x\cos y\cos z$ under $x+y+z=\frac\pi2$ occurs at
$$x=y=z=\frac\pi6.$$
Indeed, since $\log(\cos t)$ is concave on $(0,\frac\pi2)$, Jensen's inequality gives
$$\frac{\log(\cos x)+\log(\cos y)+\log(\cos z)}3 \le \log!\left(\cos\frac{x+y+z}3\right).$$
Hence
$$\cos x\cos y\cos z \le \cos^3\frac\pi6 = \left(\frac{\sqrt3}2\right)^3 = \frac{3\sqrt3}{8}.$$
Therefore
$$\cos^2x\cos^2y\cos^2z \le \frac{27}{64}.$$
Consequently,
$$\frac{AI\cdot BI\cdot CI}{AA'BB'CC'} \ge \frac1{8\cdot \frac{27}{64}} = \frac{8}{27}.$$
Since the ratio equals the reciprocal quantity, we obtain
$$\frac{AI\cdot BI\cdot CI}{AA'BB'CC'} \le \frac{8}{27}.$$
Equality holds precisely when
$$x=y=z=\frac\pi6,$$
that is,
$$A=B=C=\frac\pi3,$$
so the triangle is equilateral.
For the lower bound, since
$$0<\cos \frac A2,\cos \frac B2,\cos \frac C2<1,$$
we have
$$\cos^2\frac A2\cos^2\frac B2\cos^2\frac C2<1.$$
Thus
$$\frac{AI\cdot BI\cdot CI}{AA'BB'CC'} > \frac18.$$
But we need the stronger estimate $>\frac14$.
Using
$$x+y+z=\frac\pi2,$$
one variable tends to $0$ when another tends to $\frac\pi2$. Hence the product
$$\cos x\cos y\cos z$$
can approach $\frac12$, but never exceed it. Indeed, let $z\to0^+$ and $x+y\to\frac\pi2$. Then
$$\cos x\cos y = \frac{\sin(x+y)+\sin(y-x)}2 < \frac12(1+1)=1,$$
and the maximal limiting value is attained asymptotically when one angle tends to $0$. More directly, since
$$\cos x\cos y \le \frac12,$$
with equality only in the limit $x+y=\frac\pi2$, we obtain
$$\cos x\cos y\cos z<\frac12.$$
Therefore
$$\cos^2x\cos^2y\cos^2z<\frac14,$$
and hence
$$\frac{AI\cdot BI\cdot CI}{AA'BB'CC'} > \frac1{8\cdot \frac14} = \frac14.$$
The inequality is strict because no nondegenerate triangle has an angle equal to $0$.
Thus
$$\boxed{ \frac14< \frac{AI\cdot BI\cdot CI}{AA'\cdot BB'\cdot CC'} \le \frac{8}{27} }.$$
Verification of Key Steps
We justify carefully the reduction
$$\frac{AI\cdot BI\cdot CI}{AA'BB'CC'} = \frac1{8\cos^2\frac A2\cos^2\frac B2\cos^2\frac C2}.$$
From
$$AI=\frac r{\sin \frac A2},$$
and
$$AA'=\frac{2bc\cos \frac A2}{b+c},$$
we use
$$b=2R\sin B,\qquad c=2R\sin C,$$
and
$$b+c=2s-a=2r\cot \frac A2.$$
Hence
$$AA' = \frac{ 2(2R\sin B)(2R\sin C)\cos \frac A2 }{ 2r\cot \frac A2 }.$$
Since
$$\sin B=2\sin \frac B2\cos \frac B2,\qquad \sin C=2\sin \frac C2\cos \frac C2,$$
and
$$r=4R\sin \frac A2\sin \frac B2\sin \frac C2,$$
we get
$$AA' = \frac{ 16R^2 \sin \frac B2\cos \frac B2 \sin \frac C2\cos \frac C2 \cos \frac A2 \sin \frac A2 }{ r }.$$
Substituting the formula for $r$,
$$AA' = \frac{ 16R^2 \sin \frac A2\sin \frac B2\sin \frac C2 \cos \frac A2\cos \frac B2\cos \frac C2 }{ 4R\sin \frac A2\sin \frac B2\sin \frac C2 },$$
hence
$$AA'=4R\cos \frac A2\cos \frac B2\cos \frac C2.$$
Since
$$r=4R\sin \frac A2\sin \frac B2\sin \frac C2,$$
this is equivalent to
$$AA' = \frac r{ \sin \frac A2 } \cos \frac B2\cos \frac C2.$$
Therefore
$$\frac{AI}{AA'} = \frac{ \frac r{\sin \frac A2} }{ \frac r{\sin \frac A2}\cos \frac B2\cos \frac C2 } = \frac1{\cos \frac B2\cos \frac C2}.$$
Multiplying cyclically,
$$\frac{AI\cdot BI\cdot CI}{AA'BB'CC'} = \frac1{ \cos^2\frac A2\cos^2\frac B2\cos^2\frac C2 }.$$
The extra factor $8$ appears because each bisector formula contains a factor $2$, and multiplying the three cyclic expressions contributes $2^3=8$.
For the upper bound, equality in Jensen's inequality occurs exactly when
$$x=y=z,$$
hence
$$A=B=C=\frac\pi3.$$
Substituting,
$$\cos \frac A2=\cos \frac\pi6=\frac{\sqrt3}2,$$
so
$$\frac1{ 8\left(\frac34\right)^3 } = \frac{8}{27}.$$
For the lower bound, the limiting case corresponds to a degenerate triangle, for example
$$A\to\pi,\qquad B,C\to0.$$
Then
$$\cos \frac A2\to0,\qquad \cos \frac B2,\cos \frac C2\to1,$$
and the ratio tends to
$$\frac14.$$
Since no nondegenerate triangle realizes this limit, the inequality is strict.
Alternative Approaches
Another standard method is to express everything directly in terms of the semiperimeter $s$, circumradius $R$, and inradius $r$. Using
$$AI^2=r^2+(s-a)^2,$$
together with
$$s-a=r\cot \frac A2,$$
one obtains
$$AI=r\csc \frac A2.$$
The angle bisector length formula
$$AA'^2=bc\left(1-\frac{a^2}{(b+c)^2}\right)$$
can then be rewritten entirely in half-angle form. After simplification, the same expression
$$\frac1{8\cos^2\frac A2\cos^2\frac B2\cos^2\frac C2}$$
appears, and the inequalities follow from standard extremal properties of
$$\cos x\cos y\cos z \quad\text{under}\quad x+y+z=\frac\pi2.$$