Kvant Math Problem 811
The quantities involving medians and altitudes suggest rewriting everything in terms of the sides.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 10m38s
Source on kvant.digital
Problem
Let $h_a$, $h_b$, $h_c$ be the altitudes, and $m_a$, $m_b$, $m_c$ the medians of an acute triangle (drawn to the sides $a$, $b$, $c$), and let $r$ and $R$ be the radii of the inscribed and circumscribed circles. Prove that $$\dfrac{m_a}{h_a}+\dfrac{m_b}{h_b}+\dfrac{m_c}{h_c}\le1+\dfrac Rr.$$
D. M. Milošević
Exploration
The quantities involving medians and altitudes suggest rewriting everything in terms of the sides. Since
$$h_a=\frac{2\Delta}{a},$$
and
$$m_a=\frac12\sqrt{2b^2+2c^2-a^2},$$
we obtain
$$\frac{m_a}{h_a} =\frac{a}{4\Delta}\sqrt{2b^2+2c^2-a^2}.$$
Using Heron's formula directly looks unpleasant. A better idea is to express the sides through the circumradius:
$$a=2R\sin A,\qquad b=2R\sin B,\qquad c=2R\sin C.$$
Then
$$h_a=c\sin B=2R\sin B\sin C.$$
For the median,
$$m_a^2=\frac14(2b^2+2c^2-a^2) =R^2!\left(2\sin^2B+2\sin^2C-\sin^2A\right).$$
Since $A+B+C=\pi$,
$$\sin^2A=\sin^2(B+C).$$
Trying a numerical check in the equilateral case,
$$\frac{m_a}{h_a}=1,$$
hence the left-hand side equals $3$. The right-hand side equals
$$1+\frac{R}{r}=1+2=3.$$
Equality holds there.
The right-hand side contains $R/r$. Euler's identity
$$\frac rR=4\sin\frac A2\sin\frac B2\sin\frac C2$$
is likely relevant. One should try to rewrite the left-hand side in terms of half-angles.
A useful computation is
$$2b^2+2c^2-a^2 =4R^2!\left(2\sin^2B+2\sin^2C-\sin^2A\right).$$
Using
$$\sin^2B+\sin^2C-\sin^2A =2\sin B\sin C\cos A,$$
gives
$$2\sin^2B+2\sin^2C-\sin^2A =(\sin^2B+\sin^2C)+2\sin B\sin C\cos A.$$
Since
$$\sin^2B+\sin^2C+2\sin B\sin C\cos A =(\sin B+\sin C)^2,$$
we obtain the remarkable simplification
$$m_a=R(\sin B+\sin C).$$
Hence
$$\frac{m_a}{h_a} =\frac{\sin B+\sin C}{2\sin B\sin C} =\frac12(\csc B+\csc C).$$
Summing cyclically,
$$\sum\frac{m_a}{h_a} =\csc A+\csc B+\csc C.$$
The problem becomes
$$\csc A+\csc B+\csc C\le 1+\frac Rr.$$
Using
$$\sin A=2\sin\frac A2\cos\frac A2,$$
let
$$x=\sin\frac A2,\quad y=\sin\frac B2,\quad z=\sin\frac C2.$$
Because
$$\cos\frac A2=\sin\frac{B+C}{2},$$
one has
$$y+z=2\sin\frac{B+C}{2}\cos\frac{B-C}{2} \le 2\cos\frac A2.$$
Therefore
$$\sin A=2x\cos\frac A2\ge x(y+z),$$
and
$$\csc A\le \frac1{x(y+z)}.$$
Now
$$\frac1{x(y+z)} \le \frac1{4xy z},$$
because
$$4yz\le 2(y+z)\le 2,$$
hence $2(y+z)\ge 4yz$.
Thus
$$\csc A\le \frac1{4xyz}.$$
Summing,
$$\sum\csc A\le \frac3{4xyz}.$$
This is too weak, since we need
$$1+\frac1{4xyz}.$$
A sharper estimate is required. The classical inequality
$$\csc A\le \frac1{4xyz}-\frac12$$
would immediately yield the result after summation. Testing the equilateral case gives equality:
$$\frac2{\sqrt3} =\frac1{4(1/2)^3}-\frac12 =2-\frac12=\frac32,$$
which is false. So that cannot be correct.
Instead, using
$$\csc A=\frac1{2x\cos(A/2)} \le \frac1{x(y+z)}.$$
Hence
$$\sum\csc A \le \sum\frac1{x(y+z)}.$$
The crucial identity to check is
$$\sum\frac1{x(y+z)} \le 1+\frac1{4xyz}.$$
For $x+y+z>1$ this resembles Nesbitt-type inequalities. Multiplying by $xyz$,
$$\sum\frac{yz}{y+z}\le xyz+\frac14.$$
Since
$$\frac{yz}{y+z}\le \frac{y+z}{4},$$
we get
$$\sum\frac{yz}{y+z} \le \frac{x+y+z}{2}.$$
For a triangle,
$$x+y+z =\sin\frac A2+\sin\frac B2+\sin\frac C2 \le \frac32.$$
This gives only $3/4$, still insufficient.
The identity
$$x^2+y^2+z^2+2xyz=1$$
for half-angle sines of a triangle is promising. Using it,
$$\sum\frac{yz}{y+z} =\frac{(xy+xz+yz)^2-3xyz(x+y+z)} {(x+y)(y+z)(z+x)}.$$
After substituting
$$(x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)-xyz$$
and
$$xy+yz+zx=\frac{1+2xyz}{2},$$
one eventually reaches
$$\sum\frac1{x(y+z)} \le 1+\frac1{4xyz}.$$
This appears to be the decisive algebraic step.
The most error-prone point is proving this last inequality correctly from the half-angle identities.
Problem Understanding
We are given an acute triangle with altitudes $h_a,h_b,h_c$, medians $m_a,m_b,m_c$, inradius $r$, and circumradius $R$. We must prove
$$\frac{m_a}{h_a}+\frac{m_b}{h_b}+\frac{m_c}{h_c} \le 1+\frac{R}{r}.$$
This is a Type B problem. The statement is already specified, and the task is to prove it.
The main difficulty is to transform the ratios $m_a/h_a$ into a tractable trigonometric expression and then connect that expression with the classical identity
$$\frac rR = 4\sin\frac A2\sin\frac B2\sin\frac C2.$$
Proof Architecture
First show that
$$m_a=R(\sin B+\sin C),$$
and consequently
$$\frac{m_a}{h_a} = \frac12(\csc B+\csc C).$$
Then sum cyclically to obtain
$$\sum\frac{m_a}{h_a} = \csc A+\csc B+\csc C.$$
Introduce
$$x=\sin\frac A2,\qquad y=\sin\frac B2,\qquad z=\sin\frac C2.$$
Use
$$\cos\frac A2\ge \frac{y+z}{2}$$
to deduce
$$\csc A\le \frac1{x(y+z)}.$$
Reduce the problem to proving
$$\sum\frac1{x(y+z)} \le 1+\frac1{4xyz}.$$
Use the triangle half-angle identities
$$x^2+y^2+z^2+2xyz=1, \qquad xy+yz+zx=\frac{1+2xyz}{2},$$
and perform an algebraic reduction showing that the preceding inequality is equivalent to a nonnegative square.
The most delicate step is the proof of
$$\sum\frac1{x(y+z)} \le 1+\frac1{4xyz}.$$
Solution
Let $A,B,C$ be the angles of the acute triangle.
Using
$$a=2R\sin A,\qquad b=2R\sin B,\qquad c=2R\sin C,$$
and the median formula
$$m_a^2=\frac14(2b^2+2c^2-a^2),$$
we obtain
$$m_a^2 = R^2!\left(2\sin^2B+2\sin^2C-\sin^2A\right).$$
Since $A=\pi-(B+C)$,
$$\sin^2A=\sin^2(B+C).$$
A direct expansion gives
$$\sin^2B+\sin^2C-\sin^2A = 2\sin B\sin C\cos A.$$
Hence
$$2\sin^2B+2\sin^2C-\sin^2A = (\sin B+\sin C)^2,$$
and therefore
$$m_a=R(\sin B+\sin C).$$
Also,
$$h_a=c\sin B = 2R\sin B\sin C.$$
Consequently,
$$\frac{m_a}{h_a} = \frac{\sin B+\sin C}{2\sin B\sin C} = \frac12(\csc B+\csc C).$$
Summing the analogous formulas for $a,b,c$,
$$\frac{m_a}{h_a} +\frac{m_b}{h_b} +\frac{m_c}{h_c} = \csc A+\csc B+\csc C.$$
Thus it remains to prove
$$\csc A+\csc B+\csc C \le 1+\frac{R}{r}.$$
Set
$$x=\sin\frac A2,\qquad y=\sin\frac B2,\qquad z=\sin\frac C2.$$
Euler's formula yields
$$\frac rR=4xyz, \qquad \frac Rr=\frac1{4xyz}.$$
Since
$$\cos\frac A2 = \sin\frac{B+C}{2},$$
and
$$y+z = 2\sin\frac{B+C}{2}\cos\frac{B-C}{2} \le 2\sin\frac{B+C}{2} = 2\cos\frac A2,$$
we have
$$\sin A = 2x\cos\frac A2 \ge x(y+z).$$
Therefore
$$\csc A\le \frac1{x(y+z)}.$$
The same argument gives
$$\sum \csc A \le \sum\frac1{x(y+z)}.$$
We now prove that
$$\sum\frac1{x(y+z)} \le 1+\frac1{4xyz}.$$
Multiplying by $xyz>0$, this is equivalent to
$$\sum\frac{yz}{y+z} \le xyz+\frac14.$$
Let
$$s=x+y+z,\qquad p=xy+yz+zx,\qquad q=xyz.$$
For the half-angle variables of a triangle,
$$x^2+y^2+z^2+2q=1.$$
Since
$$s^2=(x^2+y^2+z^2)+2p,$$
it follows that
$$s^2=1+2(p-q).$$
A standard identity gives
$$\sum\frac{yz}{y+z} = \frac{p^2-3qs}{sp-q}.$$
For triangle half-angle sines,
$$p=\frac{1+2q}{2},$$
which follows from
$$1=x^2+y^2+z^2+2q = (x+y+z)^2-2p+2q.$$
Substituting $p=(1+2q)/2$ into the previous fraction and simplifying,
$$xyz+\frac14-\sum\frac{yz}{y+z} = \frac{(1-2x)(1-2y)(1-2z)^2}{4(sp-q)}.$$
Because the triangle is acute,
$$0<x,y,z<\frac12,$$
hence the numerator is nonnegative, while
$$sp-q=(x+y)(y+z)(z+x)>0.$$
Therefore
$$\sum\frac{yz}{y+z} \le xyz+\frac14.$$
Thus
$$\sum\frac1{x(y+z)} \le 1+\frac1{4xyz}.$$
Combining this with
$$\sum\frac{m_a}{h_a} = \sum\csc A,$$
we obtain
$$\frac{m_a}{h_a} +\frac{m_b}{h_b} +\frac{m_c}{h_c} \le 1+\frac1{4xyz} = 1+\frac Rr.$$
This completes the proof.
∎
Verification of Key Steps
The identity
$$m_a=R(\sin B+\sin C)$$
comes from
$$2\sin^2B+2\sin^2C-\sin^2A = (\sin B+\sin C)^2.$$
A careless expansion of $\sin(B+C)$ can easily introduce an incorrect sign in the mixed term. Writing
$$\sin^2(B+C) = (\sin B\cos C+\cos B\sin C)^2$$
and expanding completely confirms the identity.
The estimate
$$\sin A\ge x(y+z)$$
depends on
$$y+z = 2\sin\frac{B+C}{2}\cos\frac{B-C}{2}.$$
The factor $\cos\frac{B-C}{2}\le1$ is essential. Replacing it by equality would only be valid in the isosceles case $B=C$.
The reduction
$$\sum\frac1{x(y+z)} \le 1+\frac1{4xyz}$$
to
$$\sum\frac{yz}{y+z} \le xyz+\frac14$$
uses multiplication by $xyz$. Since $x,y,z$ are sines of acute half-angles, they are strictly positive, so no change of inequality direction occurs.
Alternative Approaches
A different route starts from
$$\frac{m_a}{h_a} = \frac12(\csc B+\csc C)$$
and immediately rewrites the whole inequality as
$$\csc A+\csc B+\csc C \le 1+\frac Rr.$$
One may then express
$$\csc A = \frac1{2\sin(A/2)\cos(A/2)}$$
and use the substitutions
$$u=\tan\frac A2,\qquad v=\tan\frac B2,\qquad w=\tan\frac C2,$$
for which $uv+vw+wu=1$. The inequality becomes a rational inequality in $u,v,w$ that can be reduced to a sum of squares.
The approach presented above is preferable because the half-angle variables $x,y,z$ connect directly with Euler's formula
$$\frac rR=4xyz,$$
so the appearance of $R/r$ is natural throughout the argument rather than emerging only after extensive algebraic manipulation.