For a triangle $ ABC,$ let $ k$ be its circumcircle with radius $ r.$ The bisectors of the inner angles $ A, B,$ and $ C$ of the triangle intersect respectively the circle $ k$ again at points $ A', B',$ and $ C'.$ Prove the inequality \[ 16Q^3 \geq 27 r^4 P,\] where $ Q$ and $ P$ are the areas of the triangles $ A'B'C'$ and $ABC$ respectively.
Problem
Source: IMO Shortlist 1989, Problem 6, ILL 14
Tags: geometry, circumcircle, geometric inequality, area of a triangle, IMO Shortlist
Sergic Primazon
20.09.2008 00:05
orl wrote: For a triangle $ ABC,$ let $ k$ be its circumcircle with radius $ r.$ The bisectors of the inner angles $ A, B,$ and $ C$ of the triangle intersect respectively the circle $ k$ again at points $ A', B',$ and $ C'.$ Prove the inequality \[ 16Q^3 \geq 27 r^4 P, \] where $ Q$ and $ P$ are the areas of the triangles $ A'B'C'$ and ABC respectively. Let : $ \angle A=2\alpha$ , $ \angle B=2\beta$ , $ \angle C=2\gamma$ , $ R$- circumcircle radius , $ r$ - incircle radius , $ s$ - semiperimeter $ \triangle ABC$ $ \angle A'C'B' = \alpha+\beta$ , $ \angle C'B'A' = \gamma+\alpha$ , $ \angle B'A'C' = \beta+\alpha$; \[ Q=2R^2sin(\alpha+\beta)sin(\gamma+\alpha)sin(\beta+\alpha)= 2R^2cos(\alpha)cos(\beta)cos(\gamma)=\frac{sR}{2};\] So : $ 16Q^3= 2s^3R^3\ge 27 R^4(sr)= 27R^4P$ ; ($ s^2 \ge \frac{27}{2} Rr$, $ P= sr$)
ASweatyAsianBoie
30.04.2021 00:49
Setting the Circumcenter as $O$, we get $P=[ABC]=[AOB]+[BOC]+[COA]=\frac{r^2}{2}(\sin{2A}+\sin{2B}+\sin{2C})$. Now, $Q=[A'B'C']=\frac{r^2}{2}[\sin{(A+B)}+\sin{(B+C)}+\sin{(C+A)}]$. We have: $$2[\sin{(A+B)}+\sin{(B+C)}+\sin{(C+A)}]^3\geq \frac{27}{2}(\sin{2A}+\sin{2B}+\sin{2C})$$By AM-GM: $$2[\sin{(A+B)}+\sin{(B+C)}+\sin{(C+A)}]^3\geq 2(\sin{(A+B)} \sin{(B+C)} \sin{(C+A)})$$$$\dots \geq 27(2\sin{(A+B)} \sin{(B+C)})(\sin{(C+A)})$$$$\dots \geq 27(\cos{(A-C)}-\cos{(A+2B+C)})(\sin{(C+A)})$$$$\dots \geq 27(\frac{1}{2})((\sin{2A}+\sin{2C})-(\sin{(2A+2B+2C)}+ \sin{(-2C)})$$$$\dots \geq \frac{27}{2}(\sin{2A}+\sin{2B}+\sin{2C})$$which is our original expression. $\blacksquare$
pinkpig
03.08.2021 23:54
Let $AB=c,BC=a,AC=b,A'B'=c',B'C'=a',A'C'=b',$ and $s=\frac{a+b+c}{2}.$ Now, note that $$\angle{C'B'A'}=\frac{\stackrel{\mbox{\large$\frown$}}{A'B}+\stackrel{\mbox{\large$\frown$}}{C'B}}{2}=\frac{\angle{C}+\angle{B}}{2}=\frac{180-\angle{B}}{2}=90-\frac{\angle{B}}{2}\Leftrightarrow$$$$\sin{\angle{C'B'A'}=\sin{90-\frac{\angle{B}}{2}}}=\cos{\frac{\angle{B}}{2}}=\sqrt{\frac{(s)(s-b)}{ab}}.$$Also, from the Extended Law of Sines, $$\frac{b'}{\cos{\frac{\angle{B}}{2}}}=2r\Leftrightarrow b'=2r\cos{\frac{\angle{B}}{2}}$$Now, we can use the Sine Area Formula to find that $$Q^3=\frac{\left(a'c'\cos{\frac{\angle{B}}{2}}\right)\left(a'b'\cos{\frac{\angle{C}}{2}}\right)\left(b'c'\cos{\frac{\angle{A}}{2}}\right)}{2^3}=$$$$\frac{(a'b'c')^2\left(\cos{\frac{\angle{A}}{2}}\cos{\frac{\angle{B}}{2}}\cos{\frac{\angle{C}}{2}}\right)}{8}=$$$$\frac{64r^6\left(\cos{\frac{\angle{A}}{2}}\cos{\frac{\angle{B}}{2}}\cos{\frac{\angle{C}}{2}}\right)^3}{8}=$$$$8r^6\left(\sqrt{\frac{(s)(s-a)}{bc}}\sqrt{\frac{(s)(s-b)}{ac}}\sqrt{\frac{(s)(s-c)}{ab}}\right)^3=$$$$8r^6\left(\frac{s\sqrt{s(s-a)(s-b)(s-c)}}{abc}\right)=$$$$8r^6\left(\frac{sP}{abc}\right)^3=8r^6\left(\frac{sP}{4Pr}\right)^3=8r^6\left(\frac{s}{4r}\right)^3=\frac{r^3s^3}{8}=$$$$\frac{r^3\left(\left(\frac{3}{2}\right)\left(\frac{a+b+c}{3}\right)\right)^3}{8}\geq \frac{27}{8}r^3abc=\frac{\frac{27}{8}r^34rP}{8}=$$$$\frac{\frac{27}{2}r^4P}{8}=\frac{27}{8}r^4P.$$We conclude that $$Q^3\geq\frac{27}{16}r^4P\Leftrightarrow16Q^3\geq27r^4P.$$Equality holds if $ABC$ is regular.
Sourorange
10.12.2023 17:09
Sourorange wrote: Setting the Circumcenter as $O$, we get $P=[ABC]=[AOB]+[BOC]+[COA]=\frac{r^2}{2}(\sin{2A}+\sin{2B}+\sin{2C})$. Now, $Q=[A'B'C']=\frac{r^2}{2}[\sin{(A+B)}+\sin{(B+C)}+\sin{(C+A)}]$. We have: $$2[\sin{(A+B)}+\sin{(B+C)}+\sin{(C+A)}]^3\geq \frac{27}{2}(\sin{2A}+\sin{2B}+\sin{2C})$$By AM-GM: $$2[\sin{(A+B)}+\sin{(B+C)}+\sin{(C+A)}]^3\geq 2(\sin{(A+B)} \sin{(B+C)} \sin{(C+A)})$$$$\dots \geq 27(2\sin{(A+B)} \sin{(B+C)})(\sin{(C+A)})$$$$\dots \geq 27(\cos{(A-C)}-\cos{(A+2B+C)})(\sin{(C+A)})$$$$\dots \geq 27(\frac{1}{2})((\sin{2A}+\sin{2C})-(\sin{(2A+2B+2C)}+ \sin{(-2C)})$$$$\dots \geq \frac{27}{2}(\sin{2A}+\sin{2B}+\sin{2C})$$which is our original expression. $\blacksquare$