Acute triangle $\triangle ABC$ has area $S$ and perimeter $L$. A point $P$ inside $\triangle ABC$ has $dist(P,BC)=1, dist(P,CA)=1.5, dist(P,AB)=2$. Let $BC \cap AP = D$, $CA \cap BP = E$, $AB \cap CP= F$. Let $T$ be the area of $\triangle DEF$. Prove the following inequality. $$ \left( \frac{AD \cdot BE \cdot CF}{T} \right)^2 > 4L^2 + \left( \frac{AB \cdot BC \cdot CA}{24S} \right)^2 $$
Problem
Source: 2016 KMO Senior #3
Tags: geometry, geometric inequality, Korea
12.11.2016 12:01
Yes.. This was very weird.
12.11.2016 12:03
My friend apparently Lagrange Multiplier'd this one.
12.11.2016 12:06
rkm0959 wrote: My friend apparently Lagrange Multiplier'd this one. Solution using Lagrange Multiplier??
07.01.2017 06:11
Anyone has a solution of this problem??
25.07.2017 12:45
Let the perpendicular pedal on $BC$, $AB$ from $P$ be $H$, $I$, respectively. Let $\alpha$ be the acute angle between $AC$, $PI$ and let $\beta$ be the acute angle between $AC$, $PH$. From the condition, we have $0<\alpha$, $0< \beta$, $\alpha+\beta < \frac{\pi}{2}$. \begin{eqnarray*} && \frac{\overline{AD} \cdot \overline{BE} \cdot \overline{CF}}{T}\\ &=& \frac{1}{24} (\sec \alpha) (\sec \beta)(2 (\cos (\alpha)+2 \cos (\beta)) \csc (\alpha+\beta)+3)\\ && \times \sqrt{(24 \sin (\alpha)+25) (12 \sin (\beta)+13) (4 \cos (\alpha+\beta)+5)}\\ \end{eqnarray*}\begin{eqnarray*} &&\frac{\overline{AB} \cdot \overline{BC} \cdot \overline{CA}}{24S}\\ &=& \frac{1}{24} \csc (\alpha+\beta) (3 (\tan (\alpha)+\tan (\beta))+4 \sec (\alpha)+2 \sec (\beta))\\ \end{eqnarray*}\begin{eqnarray*} && 2L \\ &=& 3 (\tan \alpha+\tan \beta)+7 \sec (\alpha)+5 \sec (\beta)\\ && +\csc (\alpha+\beta) (4 \sec (\alpha) \cos (\beta)+2 \cos (\alpha) \sec (\beta)+6) \end{eqnarray*}Therefore \begin{eqnarray*} &&144\left(\left(\frac{\overline{AD} \cdot \overline{BE} \cdot \overline{CF}}{T}\right)^2 -\left(4L^2 + \left(\frac{\overline{AB} \cdot \overline{BC} \cdot \overline{CA}}{24S}\right)^2\right)\right)\\ &=& (12 \sin (2 \alpha+\beta)+6 \sin (\alpha+2 \beta)+36 \cos (\alpha-\beta)+\cos (\alpha+\beta)+96 \sin (\alpha)+75 \sin (\beta)+118)\\ && \times (3 \sec (\alpha) \sec (\beta)+2 \csc (\alpha+\beta) (2 \sec (\alpha)+\sec (\beta)))^2 \end{eqnarray*}which is positive.
25.07.2017 16:36
Let $a=y+z,b=x+z,c=x+y$ be the length of $BC$, $CA$, $AB$, respectively. Using ratio of areas of triangles, we have $BD: CD = 4c:3b$, $CE:AE= a:2c$, $AF:BF = 3b:2a$. Therefore, let $m=BD$, $n=DC$, $p=CE$, $q=EA$, $r=AF$, $s=FB$ to write $m = \frac{4ac}{4c+3b}$, $n=\frac{3ba}{4c+3b}$, $p=\frac{ab}{a+2c}$, $q=\frac{2bc}{a+2c}$, $r=\frac{3bc}{3b+2a}$, $s=\frac{2ac}{3b+2a}$. By the Stewart's theorem, \begin{eqnarray*} AD^2 &=& \frac{b c \left(-12 a^2+12 b^2+25 b c+12 c^2\right)}{(3 b+4 c)^2}\\ BE^2 &=& \frac{a c \left(2 a^2+5 a c-2 b^2+2 c^2\right)}{(a+2 c)^2}\\ CF^2 &=& \frac{a b \left(6 a^2+13 a b+6 b^2-6 c^2\right)}{(2 a+3 b)^2}\\ \end{eqnarray*}Again, by the ratio analysis, $$ \frac{T}{S} = 1-\frac{np}{ab}-\frac{qr}{bc}-\frac{ms}{ac}= \frac{24 a b c}{(2 a+3 b) (a+2 c) (3 b+4 c)} $$On the other hand, by multiplying $24^2S^2$ on the both sides of the questioned inequality, we have the equivalent $$ \frac{24^2AD^2 \cdot BE^2 \cdot {CF}^2}{\frac{T^2}{S^2}} > 4\cdot 24^2 S^2 L^2 + a^2b^2c^2 $$Here, let $lhs=\frac{24^2{AD}^2 \cdot {BE}^2 \cdot {CF}^2}{\frac{T^2}{S^2}}$. Then we have $$ lhs = \left(49x(x+y+z)+y z\right) \left(9 y(x+y+z) +x z\right) \left(25z(x+y+z)+x y\right) $$Again, noting that $S^2= xyz(x+y+z)$, the right hand side $rhs=4\cdot 24^2 S^2 L^2 + a^2b^2c^2$ becomes $$ rhs=9216 x y z (x+y+z)^3+(x+y)^2 (x+z)^2 (y+z)^2 $$Therefore, \begin{eqnarray*} && lhs-rhs\\ &=& 440 x^4 y^2+1856 x^4 y z+1224 x^4 z^2+880 x^3 y^3\\ && +6352 x^3 y^2 z+7920 x^3 y z^2+2448 x^3 z^3+440 x^2 y^4\\ &&+6312 x^2 y^3 z+12736 x^2 y^2 z^2+7896 x^2 y z^3+1224 x^2 z^4\\ &&+1816 x y^4 z+5880 x y^3 z^2+5896 x y^2 z^3+1832 x y z^4\\ && +224 y^4 z^2+448 y^3 z^3+224 y^2 z^4 \end{eqnarray*}which is clearly positive. (Note that observing the form of lhs and rhs, one may get the result without full expansion.)
25.09.2017 06:41
Wouldn't there be more beautiful solution? All the ways on here are too mess..
06.05.2018 03:12
@Skravin I agree. The above solutions are obtained with the help of CAS. Observing that the inequality is far from being optimal, it seems that there may exists more brilliant solution. So I'm looking forward to see a smarter solution.
08.10.2019 04:56
let $x =2a,y=3b,z=4c$ Then, $ \overline {BD} : \overline {DC} = z:y$ By the Stewart's $ \overline {AD}^2 = \frac {y}{y+z} c^2 + \frac {y}{y+z} c^2 - \frac {yz}{(y+z)^2} a^2 $ it is easy to show $ T = \frac{2xyz}{(x+y)(y+z)(z+x)} S $ So this ineq follows $ \prod ({ c^2 +b^2 -a^2 + \frac{y}{z} c^2 + \frac {z} {y} b^2 }) > 16S^2 L^2 + \frac{a^2 b^2 c^2 }{144}$ Since, $ \frac{y}{z}c^2 + \frac{z}{y}b^2 = 2bc + \frac{1}{12}bc $ . it's enough to show $ ((a+b)^2 -c^2 + \frac{1}{6}ab )((b+c)^2 -a^2 + \frac{1}{12} bc ) ( (c+a)^2 - b^2 + \frac{1}{2}ca) > 16S^2 L^2 + \frac{a^2 b^2 c^2 }{144} $ On the other hand, $ \prod( (a+b)^2 -c^2 ) = (a+b+c)^3(a+b-c)(a-b+c)(-a+b+c) = 16S^2 L^2 $ So it ends because $ (U+u)(V+v)(W+w) > UVW + uvw $ ( Let $U = (a+b)^2 -c^2 , u = \frac{1}{6}ab$, and so on)