Kvant Math Problem 77

Let the triangle have sides adjacent to angle $A$ equal to $10$ and $15$.

kvantmathematicsolympiad

Verified: no
Verdicts: SKIP + SKIP
Solve time: 3m23s
Source on kvant.digital

Problem

The lengths of two sides of a triangle are 10 and 15. Prove that the bisector of the angle between them is at most 12.

N. B. Vasiliev

Exploration

Let the triangle have sides adjacent to angle $A$ equal to $10$ and $15$. Let $l$ denote the length of the bisector of angle $A$.

A standard formula for the length of an internal angle bisector is

$$l=\frac{2bc\cos\frac A2}{b+c},$$

where $b=10$ and $c=15$. This gives

$$l=\frac{300}{25}\cos\frac A2=12\cos\frac A2.$$

Since $\cos\frac A2\le 1$, this immediately yields $l\le 12$.

Before accepting this, it is useful to test a few configurations. If $A=60^\circ$, then

$$l=12\cos30^\circ=6\sqrt3\approx10.39.$$

If $A=120^\circ$, then

$$l=12\cos60^\circ=6.$$

As $A$ becomes very small, $\cos(A/2)$ approaches $1$, and $l$ approaches $12$. The bound $12$ is thus plausible and appears to be best possible, although equality would require $A=0^\circ$, which is not a genuine triangle.

The only potentially delicate point is the angle bisector formula itself. The formal proof should either derive it or justify it carefully.

Problem Understanding

We are given a triangle with two sides of lengths $10$ and $15$. The angle between these sides is bisected by an internal angle bisector. We must prove that the length of this bisector does not exceed $12$.

This is a Type B problem. The statement to be proved is already specified.

The core difficulty is obtaining an expression for the bisector length that depends on the angle between the sides and then showing that this expression is bounded above by $12$.

Proof Architecture

Let $A$ be the angle between the sides of lengths $10$ and $15$, and let $AD$ be its bisector.

First, prove the formula

$$AD=\frac{2bc\cos(A/2)}{b+c}$$

for a triangle with adjacent sides $b$ and $c$. This follows from the Angle Bisector Theorem together with Stewart's theorem.

Next, substitute $b=10$ and $c=15$ to obtain

$$AD=12\cos(A/2).$$

Finally, use the inequality $\cos(A/2)\le 1$ to conclude $AD\le 12$.

The step most likely to fail under scrutiny is the derivation of the bisector-length formula.

Solution

Let the triangle be $ABC$, with

$$AB=10,\qquad AC=15.$$

Let $AD$ be the bisector of angle $A$, where $D\in BC$. Denote

$$AD=l.$$

By the Angle Bisector Theorem,

$$\frac{BD}{DC}=\frac{AB}{AC}=\frac{10}{15}=\frac23.$$

Hence there exists a positive number $t$ such that

$$BD=2t,\qquad DC=3t.$$

Since

$$BC=BD+DC=5t,$$

Stewart's theorem applied to cevian $AD$ gives

$$AB^2\cdot DC+AC^2\cdot BD = BC,(AD^2+BD\cdot DC).$$

Substituting the known quantities,

$$10^2(3t)+15^2(2t) = 5t,(l^2+6t^2).$$

After simplification,

$$300t+450t = 5t,(l^2+6t^2),$$

so

$$750t=5t,(l^2+6t^2).$$

Since $t>0$,

$$150=l^2+6t^2.$$

Thus

$$l^2=150-6t^2.$$

Now let $A$ be the angle between the sides $10$ and $15$. By the law of cosines,

$$BC^2=10^2+15^2-2\cdot10\cdot15\cos A =325-300\cos A.$$

Because $BC=5t$,

$$25t^2=325-300\cos A,$$

hence

$$t^2=13-12\cos A.$$

Substituting into the expression for $l^2$,

$$l^2 = 150-6(13-12\cos A) = 72+72\cos A.$$

Using

$$1+\cos A=2\cos^2\frac A2,$$

we obtain

$$l^2 = 144\cos^2\frac A2.$$

Since $l>0$,

$$l=12\cos\frac A2.$$

For a triangle, $0<A<\pi$, so

$$0<\frac A2<\frac\pi2,$$

and therefore

$$\cos\frac A2\le 1.$$

Consequently,

$$l=12\cos\frac A2\le 12.$$

This completes the proof.

∎

Verification of Key Steps

The derivation from Stewart's theorem deserves a separate check. The Angle Bisector Theorem gives the ratio $BD:DC=2:3$. Writing $BD=2t$ and $DC=3t$ makes $BC=5t$. Stewart's theorem then yields

$$100(3t)+225(2t)=5t(l^2+6t^2),$$

which simplifies to

$$150=l^2+6t^2.$$

No sign changes or cancellations are hidden in this computation.

The conversion from $l^2=72+72\cos A$ to the half-angle form also merits verification. Since

$$1+\cos A=2\cos^2\frac A2,$$

we have

$$72+72\cos A = 72(1+\cos A) = 144\cos^2\frac A2.$$

Taking the positive square root is legitimate because a segment length is positive.

A common mistake would be to write $l\le12$ directly from $l^2\le144$ without checking that $l$ is nonnegative. Here $l$ is a length, so $l>0$, and the implication is valid.

Alternative Approaches

A shorter solution begins with the standard angle-bisector formula

$$l=\frac{2bc\cos(A/2)}{b+c},$$

where $b$ and $c$ are the sides adjacent to the bisected angle. Substituting $b=10$ and $c=15$ gives

$$l=\frac{2\cdot10\cdot15}{10+15}\cos\frac A2 =12\cos\frac A2.$$

Since $\cos(A/2)\le1$, the inequality $l\le12$ follows immediately.

The main solution is preferable because it derives the needed formula from elementary theorems rather than assuming it as known.