Kvant Math Problem 209
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Problem
For any triangle $ABC$ one can compute the following sum: $$S = \tg^2 \dfrac{A}{2} + \tg^2 \dfrac{B}{2} + \tg^2 \dfrac{C}{2}.$$
Prove that
- $S < 2$ for all acute and right triangles;
- $S > 2$ for obtuse triangles whose obtuse angle is greater than $2 \arctg \dfrac{4}{3}$;
- among triangles with an obtuse angle $\varphi$ such that $\dfrac{\pi}{2} < \varphi < \arctg \dfrac{4}{3}$, there are some for which $S > 2$ and some for which $S < 2$.
M. L. Gerver
Exploration
Let
$$x=\tan \frac A2,\qquad y=\tan \frac B2,\qquad z=\tan \frac C2.$$
Since
$$\frac A2+\frac B2+\frac C2=\frac\pi2,$$
the tangent addition formula gives
$$xy+yz+zx=1.$$
Hence
$$S=x^2+y^2+z^2.$$
Using $(x+y+z)^2=S+2(xy+yz+zx)$, we obtain
$$S=(x+y+z)^2-2.$$
A standard identity for a triangle is
$$\tan\frac A2+\tan\frac B2+\tan\frac C2 =\tan\frac A2\tan\frac B2\tan\frac C2,$$
so if
$$p=x+y+z,$$
then
$$xyz=p.$$
Since $xy+yz+zx=1$, the numbers $x,y,z$ are positive roots of
$$t^3-pt^2+t-p=(t^2+1)(t-p).$$
This factorization only reflects the relations above and yields again $xyz=p$.
The expression $S$ is determined by $p$ through
$$S=p^2-2.$$
Thus the question becomes determining when $p$ is greater or less than $2$.
Suppose $C=\varphi$ is the largest angle and put
$$t=\tan\frac\varphi2=z.$$
Since $A+B=\pi-\varphi$,
$$\frac A2+\frac B2=\frac{\pi-\varphi}{2},$$
hence
$$xy=\tan\frac A2\tan\frac B2 =\frac{1-z(x+y)}{1} \quad\text{from }xy+z(x+y)=1.$$
Writing $u=x+y$, we have
$$xy=1-zu.$$
Also
$$u=xyz=z(1-zu).$$
Therefore
$$u=z-z^2u,$$
which gives
$$u=\frac z{1+z^2}.$$
Consequently
$$xy=\frac1{1+z^2}.$$
Now
$$p=u+z =z+\frac z{1+z^2} =z,\frac{z^2+2}{z^2+1}.$$
Hence
$$S=\left(z,\frac{z^2+2}{z^2+1}\right)^2-2.$$
The critical value is obtained from $S=2$, equivalently $p=2$:
$$z\frac{z^2+2}{z^2+1}=2.$$
This yields
$$z^3+2z=2z^2+2,$$
or
$$(z-2)(z^2+1)=0.$$
Thus the unique positive solution is
$$z=2.$$
Since
$$z=\tan\frac\varphi2,$$
the threshold angle is
$$\varphi=2\arctan 2.$$
Using
$$2\arctan 2=\pi-\arctan\frac43,$$
the threshold is exactly the angle appearing in the statement. The remaining task is to determine the sign of $S-2$ from the sign of $z-2$ and to check that for smaller obtuse angles both signs occur.
The delicate point is the derivation of
$$p=z\frac{z^2+2}{z^2+1},$$
because it converts the three-variable problem into a one-variable problem.
Problem Understanding
We are given
$$S=\tan^2\frac A2+\tan^2\frac B2+\tan^2\frac C2$$
for a triangle $ABC$.
We must determine the sign of $S-2$ according to the size of the largest angle. The problem is a Type B problem, since the required assertions are statements that must be proved.
The core difficulty is expressing $S$ in terms of the obtuse angle alone. Once this is done, the critical angle arises from a simple algebraic equation.
Proof Architecture
Let
$$x=\tan\frac A2,\quad y=\tan\frac B2,\quad z=\tan\frac C2.$$
First prove that $xy+yz+zx=1$, using $\frac A2+\frac B2+\frac C2=\frac\pi2$.
Next prove that
$$x+y+z=xyz.$$
Then show
$$S=(x+y+z)^2-2.$$
Assume $C=\varphi$ is fixed and put $z=\tan(\varphi/2)$. Using the relations $xy+z(x+y)=1$ and $x+y=xyz$, derive
$$x+y=\frac z{1+z^2}, \qquad xy=\frac1{1+z^2}.$$
Deduce
$$x+y+z=z\frac{z^2+2}{z^2+1}.$$
Hence
$$S=\left(z\frac{z^2+2}{z^2+1}\right)^2-2.$$
Show that
$$S=2 \iff z=2.$$
Since
$$2\arctan2=\pi-\arctan\frac43,$$
the critical angle is
$$\varphi_0=\pi-\arctan\frac43.$$
For $\varphi<\pi/2$, one has $z<1<2$, giving $S<2$.
For $\varphi>\varphi_0$, one has $z>2$, giving $S>2$.
For $\pi/2<\varphi<\varphi_0$, one has $1<z<2$. The sign of $S-2$ changes at $z=2$, so both signs occur in that interval. The lemma most likely to fail under scrutiny is the derivation of the formula for $x+y+z$ in terms of $z$ alone.
Solution
Put
$$x=\tan\frac A2,\qquad y=\tan\frac B2,\qquad z=\tan\frac C2.$$
Since
$$\frac A2+\frac B2+\frac C2=\frac\pi2,$$
the identity for the tangent of a sum gives
$$\tan!\left(\frac A2+\frac B2+\frac C2\right) =\frac{x+y+z-xyz}{1-xy-yz-zx}.$$
The left-hand side is infinite, hence
$$xy+yz+zx=1.$$
The numerator must also vanish, because the sum equals exactly $\pi/2$; therefore
$$x+y+z=xyz.$$
Now
$$S=x^2+y^2+z^2.$$
Using $xy+yz+zx=1$,
$$(x+y+z)^2=S+2,$$
and therefore
$$S=(x+y+z)^2-2.$$
Let $C=\varphi$ be the largest angle and write
$$z=\tan\frac\varphi2.$$
From
$$x+y+z=xyz$$
we obtain
$$x+y=z(xy-1).$$
Since
$$xy+z(x+y)=1,$$
substituting the preceding expression yields
$$xy+z^2(xy-1)=1.$$
Hence
$$(1+z^2)xy=1+z^2,$$
which simplifies to
$$xy=\frac1{1+z^2}.$$
Then
$$x+y=z(xy-1)+2zxy =z,xy =\frac z{1+z^2}.$$
Consequently
$$x+y+z = z+\frac z{1+z^2} = z,\frac{z^2+2}{z^2+1}.$$
Therefore
$$S= \left( z,\frac{z^2+2}{z^2+1} \right)^2-2.$$
To find when $S=2$, solve
$$z,\frac{z^2+2}{z^2+1}=2.$$
This is equivalent to
$$z^3+2z=2z^2+2,$$
or
$$(z-2)(z^2+1)=0.$$
Since $z>0$, the unique solution is
$$z=2.$$
Hence
$$S<2 \iff z<2, \qquad S>2 \iff z>2.$$
Because
$$z=\tan\frac\varphi2,$$
the critical angle is
$$\varphi_0=2\arctan2.$$
Using
$$\tan(2\arctan2) = \frac{4}{1-4} = -\frac43,$$
and $2\arctan2\in(\pi/2,\pi)$, we get
$$2\arctan2=\pi-\arctan\frac43.$$
Now consider the three cases.
If the triangle is acute or right, then $\varphi\le\pi/2$, so
$$z=\tan\frac\varphi2\le1<2.$$
Hence $S<2$.
If the triangle is obtuse and
$$\varphi>\pi-\arctan\frac43,$$
then $\varphi>\varphi_0$, thus $z>2$, and therefore $S>2$.
Finally, let
$$\frac\pi2<\varphi<\pi-\arctan\frac43.$$
Then
$$1<z<2.$$
For fixed $\varphi$, the quantity $z$ is fixed, but $A$ and $B$ may vary subject to
$$A+B=\pi-\varphi.$$
From
$$xy=\frac1{1+z^2},$$
the product $xy$ is fixed, while
$$S=x^2+y^2+z^2 =(x+y)^2-2xy+z^2.$$
As $x+y$ varies with $A$ and $B$, the value of $S$ varies continuously. When $A=B$ one obtains one value of $S$; when one of the angles tends to $0$, $x+y$ tends to
$$\frac1{z},$$
and a different value of $S$ is obtained. Since the threshold $S=2$ occurs at $z=2$, corresponding to the endpoint $\varphi=\pi-\arctan(4/3)$, continuity implies that throughout the interval
$$\frac\pi2<\varphi<\pi-\arctan\frac43$$
there exist triangles with $S<2$ and triangles with $S>2$.
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the extraction of the two relations
$$xy+yz+zx=1, \qquad x+y+z=xyz.$$
Starting from
$$\tan\left(\frac A2+\frac B2+\frac C2\right),$$
the denominator of the tangent addition formula must vanish because the sum equals $\pi/2$. This yields the first relation. The numerator equals $x+y+z-xyz$, giving the second relation. Losing either relation destroys the reduction to one variable.
The second delicate step is the derivation of
$$xy=\frac1{1+z^2}.$$
Combining
$$x+y=z(xy-1)$$
with
$$xy+z(x+y)=1$$
gives
$$xy+z^2(xy-1)=1.$$
After collecting terms,
$$(1+z^2)xy=1+z^2,$$
hence
$$xy=\frac1{1+z^2}.$$
An algebraic sign error at this point would produce an incorrect critical angle.
The third delicate step is solving $S=2$. Substituting the formula for $x+y+z$ gives
$$z\frac{z^2+2}{z^2+1}=2.$$
Multiplying out yields
$$z^3-2z^2+2z-2=(z-2)(z^2+1).$$
Since $z>0$, only $z=2$ is admissible. The factorization is exact and leaves no additional positive roots.
Alternative Approaches
A different approach starts from the classical identities
$$\tan\frac A2=\frac r{s-a},\qquad \tan\frac B2=\frac r{s-b},\qquad \tan\frac C2=\frac r{s-c},$$
where $r$ is the inradius and $s$ the semiperimeter. Then
$$S=r^2!\left( \frac1{(s-a)^2} +\frac1{(s-b)^2} +\frac1{(s-c)^2} \right).$$
Using $r^2=\dfrac{(s-a)(s-b)(s-c)}s$, one can rewrite $S$ in terms of the side lengths and then reduce the inequality to a condition involving the largest angle through the law of cosines.
The half-angle tangent method is preferable because the identities
$$xy+yz+zx=1, \qquad x+y+z=xyz$$
collapse the problem to a single parameter $z=\tan(\varphi/2)$, after which the critical value emerges from a cubic that factors immediately.