In the interior of $\triangle ABC$ we take the arbitrary point $P$. Prove that the following inequality holds: $$\small c(\sin\angle CAP + \sin\angle CBP) + a(\sin\angle ABP +\sin\angle ACP) + b(\sin\angle BCP + \sin\angle BAP)\\ \le a + b + c$$

I tried to draw the distances from $P$ to the sides of the triangle and write the sines from the right angled triangles, but I couldn't get anything out of it. I saw that a case of equality is when the triangle is equilateral and $P$ is its center.

Can you help me?

  • $\begingroup$ Interesting inequality. Where does it come from? $\endgroup$
    – ΑΘΩ
    Mar 14, 2019 at 13:14
  • $\begingroup$ It is from the romanian publishment “Gazeta Matematica”. $\endgroup$ Mar 14, 2019 at 16:17
  • $\begingroup$ If you would be as kind as to tell us what edition of the Gazette that might be, we might be able to look things up ;) $\endgroup$
    – ΑΘΩ
    Mar 14, 2019 at 16:37
  • $\begingroup$ It is from the edition 10/2018, the number of the problem being S:L18.249 :) $\endgroup$ Mar 14, 2019 at 16:53

1 Answer 1


To simplify derivation introduce the notation: $$ BC=a,\quad AC=b,\quad AB=c,\\ \measuredangle{CAB}=\alpha,\quad \measuredangle{ABC}=\beta,\quad \measuredangle{BCA}=\gamma;\\ \measuredangle{ABP}=\alpha_B,\quad \measuredangle{ACP}=\alpha_C,\\ \measuredangle{BAP}=\beta_A,\quad \measuredangle{BCP}=\beta_C,\\\ \measuredangle{CAP}=\gamma_A,\quad \measuredangle{CBP}=\gamma_B. $$ Then the initial inequality can be written as: $$\small \frac{a(\sin\alpha_B+\sin\alpha_C) +b(\sin\beta_A+\sin\beta_C)+c(\sin\gamma_A+\sin\gamma_B)}{a+b+c}\le1.\tag{*} $$ which can be proved as follows: $$\begin{align}\small &\scriptsize\frac{a(\sin\alpha_B+\sin\alpha_C) +b(\sin\beta_A+\sin\beta_C)+c(\sin\gamma_A+\sin\gamma_B)}{a+b+c}\\ &\scriptsize=\frac{\sin\alpha(\sin\alpha_B+\sin\alpha_C) +\sin\beta(\sin\beta_A+\sin\beta_C)+\sin\gamma(\sin\gamma_A+\sin\gamma_B)} {\sin\alpha+\sin\beta+\sin\gamma}\tag1\\ &\scriptsize=\frac{(\sin\beta\sin\beta_A+\sin\gamma\sin\gamma_A) +(\sin\alpha\sin\alpha_B+\sin\gamma\sin\gamma_B) +(\sin\alpha\sin\alpha_C+\sin\beta\sin\beta_C)} {\sin\alpha+\sin\beta+\sin\gamma}\tag2\\ &\scriptsize\le\frac{\sin\alpha+\sin\beta+\sin\gamma}{\sin\alpha+\sin\beta+\sin\gamma}=1\tag3. \end{align} $$


(1) Substitution: $$a=2R\sin\alpha, \quad b=2R\sin\beta, \quad c=2R\sin\gamma.$$

(2) Rearrangement.

(3) Transformation of sine products to cosine differences and then from cosine sums to cosine products: $$\begin{align} &\scriptsize\sin\beta\sin\beta_A+\sin\gamma\sin\gamma_A\\ &\scriptsize=\frac{\cos(\beta-\beta_A)-\cos(\beta+\beta_A)}2 +\frac{\cos(\gamma-\gamma_A)-\cos(\gamma+\gamma_A)}2\\ &\scriptsize=\frac{\cos(\beta-\beta_A)+\cos(\gamma-\gamma_A)}2 -\frac{\cos(\beta+\beta_A)+\cos(\gamma+\gamma_A)}2\\ &\scriptsize= \underbrace{\cos\frac{\beta+\gamma-\beta_A-\gamma_A}2}_{\ge0} \underbrace{\cos\frac{\beta-\gamma-\beta_A+\gamma_A}2}_{\le1} - \underbrace{\cos\frac{\beta+\gamma+\beta_A+\gamma_A}2}_{=0}\cos\frac{\beta-\gamma+\beta_A-\gamma_A}2\tag4\\ &\scriptsize\le \cos\frac{\beta+\gamma-\alpha}2=\cos\left(\frac\pi2-\alpha\right)=\sin\alpha, \end{align} $$ where we used $\beta_A+\gamma_A=\alpha$.

There may arise an interesting question when does the inequality $(\text{*})$ degenerate to equality?

It is easily seen from $(4)$ that this happens if and only if the following equations hold: $$ \begin {cases} \beta-\beta_A=\gamma-\gamma_A\\ \gamma-\gamma_B=\alpha-\alpha_B\\ \alpha-\alpha_C=\beta-\beta_C \end {cases}, $$ which together with $$ \begin {cases} \beta_A+\gamma_A=\alpha\\ \gamma_B+\alpha_B=\beta\\ \alpha_C+\beta_C=\gamma \end {cases} $$ implies: $$ \begin {cases} \beta_C=\gamma_B=\frac\pi2-\alpha\\ \gamma_A=\alpha_C=\frac\pi2-\beta\\ \alpha_B=\beta_A=\frac\pi2-\gamma \end {cases}. $$

Thus the equality holds if and only if the triangle $ABC$ is acute and $P$ is its circumcenter.


With the convention that the angle measures at a triangle vertex are signed and the direction towards the adjacent side is positive, the inequality (*) becomes valid for the whole plane. Correspondingly it degenerates to equality at the circumcenter of the triangle, regardless of its shape.

  • $\begingroup$ Thank you very much! $\endgroup$ Mar 15, 2019 at 14:58
  • $\begingroup$ @MichaelGoldberg You're welcome! $\endgroup$
    – user
    Mar 15, 2019 at 15:09
  • $\begingroup$ Great presentation! $\endgroup$
    – Allawonder
    Mar 16, 2019 at 10:28
  • $\begingroup$ @Allawonder Thank you! $\endgroup$
    – user
    Mar 16, 2019 at 10:38
  • 1
    $\begingroup$ A lovely solution! Just one remark though: the case of equality entails indeed the equations displayed in the final curly brace, however the conclusion on $P$ is not entirely correct. From $\sphericalangle{ABP}=\sphericalangle{BAP}$ one infers $|PA|=|PB|$ and the analogues, to the conclusion that $P$ is the circumcenter. The requirement that the triangle be acute is indeed valid. $\endgroup$
    – ΑΘΩ
    Mar 21, 2019 at 11:06

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