Kvant Math Problem 1490

The statement as written can only be meaningful if the second triangle has side lengths $\sin x$, $\sin y$, $\sin z$.

kvantmathematicsolympiad

Verified: yes
Verdicts: PASS + PASS
Solve time: 26m37s
Source on kvant.digital

Problem

Let $x$, $y$, $z$ be the side lengths of a triangle whose perimeter is less than $\pi$. Prove that

  1. the numbers $x$, $y$, $z$ can also serve as the side lengths of a triangle, and
  2. its area does not exceed $\dfrac18$ of the sum of $2x$, $2y$, $2z$.

V. A. Ufnarovsky

Exploration

The statement as written can only be meaningful if the second triangle has side lengths $\sin x$, $\sin y$, $\sin z$. Otherwise part 1 is tautological. The natural interpretation is therefore:

Given a triangle with side lengths $x,y,z$ and perimeter less than $\pi$, prove that $\sin x,\sin y,\sin z$ are also the side lengths of a triangle, and that the area of this new triangle does not exceed

$$\frac18(2x+2y+2z)=\frac{x+y+z}{4}.$$

The first part requires proving the triangle inequalities for $\sin x,\sin y,\sin z$. Since

$$x<y+z,\qquad y+z<\pi,$$

the identity

$$\sin y+\sin z = 2\sin\frac{y+z}{2}\cos\frac{y-z}{2}$$

looks useful. Because

$$\left|\frac{y-z}{2}\right|<\frac{y+z}{2},$$

we obtain

$$\cos\frac{y-z}{2}>\cos\frac{y+z}{2},$$

hence

$$\sin y+\sin z>\sin(y+z).$$

Since $x<y+z<\pi$, the inequality $\sin(y+z)>\sin x$ is not automatic, because $\sin t$ is not monotone on $(0,\pi)$. A better comparison is

$$\sin y+\sin z > \sin(y+z) = \sin x+2\cos\frac{x+y+z}{2}\sin\frac{y+z-x}{2}.$$

The perimeter condition gives $\frac{x+y+z}{2}<\frac{\pi}{2}$, so the cosine factor is positive. Thus $\sin(y+z)>\sin x$.

For the area estimate, a direct Heron computation seems unnecessary. Let $a=\sin x$, $b=\sin y$, $c=\sin z$. Since $\sin t\le t$ for $t>0$,

$$a+b+c\le x+y+z.$$

A standard bound for the area $S$ of a triangle with sides $a,b,c$ is Weitzenböck's inequality,

$$S\le \frac{(a+b+c)^2}{12\sqrt3},$$

with equality only for an equilateral triangle. Then

$$S\le \frac{(x+y+z)^2}{12\sqrt3}.$$

Because $x+y+z<\pi<3\sqrt3$, this implies

$$S<\frac{x+y+z}{4}.$$

The only delicate point is the proof of the triangle inequality for the sine sides.

Problem Understanding

We are given a triangle with side lengths $x,y,z$ whose perimeter satisfies

$$x+y+z<\pi.$$

We must prove two statements about the numbers

$$\sin x,\quad \sin y,\quad \sin z.$$

First, they satisfy the triangle inequalities and hence are themselves the side lengths of a triangle.

Second, if $S$ denotes the area of that new triangle, then

$$S\le \frac18(2x+2y+2z)=\frac{x+y+z}{4}.$$

This is a Type B problem. The core difficulty is establishing the triangle inequalities for $\sin x,\sin y,\sin z$ from the original inequalities for $x,y,z$ and the perimeter restriction.

Proof Architecture

The first lemma is that

$$\sin y+\sin z>\sin(y+z)$$

whenever $y,z>0$ and $y+z<\pi$; this follows from the sum-to-product identity.

The second lemma is that

$$\sin(y+z)>\sin x$$

whenever $x<y+z$ and $x+y+z<\pi$; this follows from the identity for $\sin(y+z)-\sin x$.

Combining these two lemmas yields

$$\sin x<\sin y+\sin z,$$

and the two analogous inequalities follow by symmetry.

The next lemma is Weitzenböck's inequality:

$$S\le \frac{(a+b+c)^2}{12\sqrt3}$$

for any triangle with sides $a,b,c$.

Finally, using $a=\sin x$, $b=\sin y$, $c=\sin z$ and $\sin t\le t$, we obtain

$$S\le \frac{(x+y+z)^2}{12\sqrt3}.$$

Since $x+y+z<\pi<3\sqrt3$, this gives

$$S<\frac{x+y+z}{4}.$$

The lemma most likely to fail under insufficient scrutiny is the proof that $\sin(y+z)>\sin x$, because $\sin t$ is not monotone on the whole interval $(0,\pi)$.

Solution

Let

$$a=\sin x,\qquad b=\sin y,\qquad c=\sin z.$$

We first prove that $a,b,c$ satisfy the triangle inequalities.

Using the identity

$$\sin y+\sin z = 2\sin\frac{y+z}{2}\cos\frac{y-z}{2},$$

and the inequalities

$$0<\frac{|y-z|}{2}<\frac{y+z}{2}<\frac{\pi}{2},$$

we obtain

$$\cos\frac{y-z}{2} > \cos\frac{y+z}{2}.$$

Hence

$$\sin y+\sin z > 2\sin\frac{y+z}{2}\cos\frac{y+z}{2} = \sin(y+z).$$

Next,

$$\sin(y+z)-\sin x = 2\cos\frac{x+y+z}{2}, \sin\frac{y+z-x}{2}.$$

Since $x+y+z<\pi$,

$$0<\frac{x+y+z}{2}<\frac{\pi}{2},$$

so

$$\cos\frac{x+y+z}{2}>0.$$

Since $x<y+z$,

$$\sin\frac{y+z-x}{2}>0.$$

Therefore

$$\sin(y+z)-\sin x>0,$$

that is,

$$\sin(y+z)>\sin x.$$

Combining the two inequalities gives

$$\sin y+\sin z>\sin x.$$

By symmetry,

$$\sin z+\sin x>\sin y, \qquad \sin x+\sin y>\sin z.$$

Thus $a,b,c$ are the side lengths of a triangle.

Let $S$ be the area of this triangle.

Weitzenböck's inequality states that for any triangle with side lengths $a,b,c$,

$$S\le \frac{(a+b+c)^2}{12\sqrt3}.$$

Since $\sin t\le t$ for every positive $t$,

$$a+b+c = \sin x+\sin y+\sin z \le x+y+z.$$

Hence

$$S \le \frac{(x+y+z)^2}{12\sqrt3}.$$

Because

$$x+y+z<\pi<3\sqrt3,$$

we have

$$\frac{x+y+z}{12\sqrt3}<\frac14.$$

Multiplying by the positive number $x+y+z$ yields

$$\frac{(x+y+z)^2}{12\sqrt3} < \frac{x+y+z}{4}.$$

Therefore

$$S<\frac{x+y+z}{4} = \frac18(2x+2y+2z).$$

Both required statements have been proved.

This completes the proof.

∎

Verification of Key Steps

The first delicate step is proving

$$\sin y+\sin z>\sin(y+z).$$

A careless argument might try to use monotonicity of $\sin$, which is false on $(0,\pi)$. The correct computation is

$$\sin y+\sin z = 2\sin\frac{y+z}{2}\cos\frac{y-z}{2},$$

and because

$$\left|\frac{y-z}{2}\right| < \frac{y+z}{2} < \frac{\pi}{2},$$

the cosine factor on the right is strictly larger than $\cos\frac{y+z}{2}$.

The second delicate step is proving

$$\sin(y+z)>\sin x.$$

The inequality $x<y+z$ alone is insufficient, because $\sin$ is not increasing on all of $(0,\pi)$. The identity

$$\sin(y+z)-\sin x = 2\cos\frac{x+y+z}{2}\sin\frac{y+z-x}{2}$$

uses both hypotheses. The perimeter bound makes the cosine positive, and the triangle inequality makes the sine positive.

The third delicate step is the final comparison

$$\frac{(x+y+z)^2}{12\sqrt3} \le \frac{x+y+z}{4}.$$

This is equivalent to

$$x+y+z\le 3\sqrt3.$$

The given hypothesis yields the stronger inequality

$$x+y+z<\pi<3\sqrt3.$$

Alternative Approaches

After establishing that $\sin x,\sin y,\sin z$ form a triangle, one may use Heron's formula together with the substitutions

$$s=\frac{x+y+z}{2},\qquad u=s-x,\quad v=s-y,\quad w=s-z.$$

A sequence of trigonometric transformations expresses the area in terms of products of sines involving $s,u,v,w$. The inequality

$$\sin t\le t$$

and the condition $s<\frac{\pi}{2}$ then lead to the required estimate.

The approach using Weitzenböck's inequality is preferable because it separates the problem into two independent parts. Once the triangle inequalities for $\sin x,\sin y,\sin z$ are established, the area bound follows immediately from a standard extremal estimate and the elementary inequality $\sin t\le t$.