Subfield: Geometry


Question: In $\triangle A B C, B C=a, A C=b, A B=c$, and $a<\frac{1}{2}(b+c)$.

Prove that $\angle B A C<\frac{1}{2}(\angle A B C+\angle A C B)$.

Solution:
We use the notation $A=\angle B A C, B=\angle A B C$ and $C=\angle A C B$.

We need to show that $A<\frac{1}{2}(B+C)$. Since the sum of the angles in $\triangle A B C$ is $180^{\circ}$, then $B+C=180^{\circ}-A$, and so this inequality is equivalent to $A<\frac{1}{2}\left(180^{\circ}-A\right)$ which is equivalent to $\frac{3}{2} A<90^{\circ}$ or $A<60^{\circ}$.

So we need to show that $A<60^{\circ}$.

We know that $a<\frac{1}{2}(b+c)$. Thus, $2 a<b+c$ and so $4 a^{2}<b^{2}+c^{2}+2 b c$ because all quantities are positive.

Using the cosine law in $\triangle A B C$, we obtain $a^{2}=b^{2}+c^{2}-2 b c \cos A$.

Therefore,

$$
\begin{aligned}
4 a^{2} & <b^{2}+c^{2}+2 b c \\
4\left(b^{2}+c^{2}-2 b c \cos A\right) & <b^{2}+c^{2}+2 b c \\
4 b^{2}+4 c^{2}-8 b c \cos A & <b^{2}+c^{2}+2 b c \\
4 b^{2}+4 c^{2}-8 b c \cos A & <b^{2}+c^{2}+2 b c+3(b-c)^{2} \quad\left(\text { since }(b-c)^{2} \geq 0\right) \\
4 b^{2}+4 c^{2}-8 b c \cos A & <b^{2}+c^{2}+2 b c+3 b^{2}-6 b c+3 c^{2} \\
4 b^{2}+4 c^{2}-8 b c \cos A & <4 b^{2}+4 c^{2}-4 b c \\
-8 b c \cos A & <-4 b c \\
\cos A & >\frac{1}{2} \quad(\text { since } 8 b c>0)
\end{aligned}
$$

Since $2 a<b+c$, then $a$ cannot be the longest side of $\triangle A B C$ (that is, we cannot have $a \geq b$ and $a \geq c$ ), so $A$ must be an acute angle.

Therefore, $\cos A>\frac{1}{2}$ implies $A<60^{\circ}$, as required.
We use the notation $A=\angle B A C, B=\angle A B C$ and $C=\angle A C B$.

We need to show that $A<\frac{1}{2}(B+C)$. Since the sum of the angles in $\triangle A B C$ is $180^{\circ}$, then $B+C=180^{\circ}-A$, and so this inequality is equivalent to $A<\frac{1}{2}\left(180^{\circ}-A\right)$ which is equivalent to $\frac{3}{2} A<90^{\circ}$ or $A<60^{\circ}$.

So we need to show that $A<60^{\circ}$.

We know that $a<\frac{1}{2}(b+c)$ which implies $2 a<b+c$.

Using the sine law in $\triangle A B C$, we obtain $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$, which gives $b=\frac{a \sin B}{\sin A}$ and $c=\frac{a \sin C}{\sin A}$.

Therefore, we obtain equivalent inequalities

$$
\begin{aligned}
2 a & <b+c \\
2 a & <\frac{a \sin B}{\sin A}+\frac{a \sin C}{\sin A} \\
2 a \sin A & <a \sin B+a \sin C \quad\left(\text { since } \sin A>0 \text { for } 0^{\circ}<A<180^{\circ}\right) \\
2 \sin A & <\sin B+\sin C
\end{aligned}
$$

since $a>0$. Next, we use the trigonometric formula $\sin B+\sin C=2 \sin \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right)$.

Since $\cos \theta \leq 1$ for any $\theta$, then $\sin B+\sin C \leq 2 \sin \left(\frac{B+C}{2}\right) \cdot 1=2 \sin \left(\frac{B+C}{2}\right)$.



Therefore,

$$
\begin{aligned}
2 \sin A & <\sin B+\sin C \leq 2 \sin \left(\frac{B+C}{2}\right) \\
2 \sin A & <2 \sin \left(\frac{B+C}{2}\right) \\
2 \sin A & <2 \sin \left(\frac{180^{\circ}-A}{2}\right) \\
4 \sin \left(\frac{1}{2} A\right) \cos \left(\frac{1}{2} A\right) & <2 \sin \left(90^{\circ}-\frac{1}{2} A\right) \\
2 \sin \left(\frac{1}{2} A\right) \cos \left(\frac{1}{2} A\right) & <\cos \left(\frac{1}{2} A\right)
\end{aligned}
$$

Since $0^{\circ}<A<180^{\circ}$, then $\cos \left(\frac{1}{2} A\right)>0$, so $\sin \left(\frac{1}{2} A\right)<\frac{1}{2}$.

Since $2 a<b+c$, then $a$ cannot be the longest side of $\triangle A B C$, so $A$ must be an acute angle.

Therefore, $\frac{1}{2} A<30^{\circ}$ or $A<60^{\circ}$, as required.