Geometric Inequality

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Source: Iranian National Olympiad (3rd Round) 2006

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1275 posts

#1 h Sep 21, 2006, 5:22 PM • 2 Y

Y by Adventure10, Mango247

In triangle $ABC$ , if $L,M,N$ are midpoints of $AB,AC,BC$ . And $H$ is orthogonal center of triangle $ABC$ , then prove that $LH^{2}+MH^{2}+NH^{2}\leq\frac14(AB^{2}+AC^{2}+BC^{2})$

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#2 h Sep 22, 2006, 8:46 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Only if the triangle $ABC$ is acute or right !

Omid Hatami wrote:

In acute or right triangle $ABC$ , if $L,M,N$ are midpoints of the its sides and $H$ is its orthocenter, then show that $\boxed{\ LH^{2}+MH^{2}+NH^{2}\leq\frac14\cdot \left(a^{2}+b^{2}+c^{2}\right)\ }$

Proof. It is well-known that $HA=2R|\cos A|$ , $a=2R\sin A$ a.s.o. and $\Delta\equiv\sum \cos 2A=-1-4\prod\cos A\ .$ Therefore,
$4\sum HL^{2}-\sum a^{2}=$ $\sum\left(4\cdot HL^{2}-a^{2}\right)=$ $\sum\left[2\left(HB^{2}+HC^{2}\right)-2a^{2}\right]=$ $4\sum HA^{2}-2\sum a^{2}=$
$16R^{2}\sum \cos^{2}A-8R^{2}\sum \sin^{2}A=$ $8R^{2}\sum (1+\cos 2A)-4R^{2}\sum (1-\cos 2A)=$ $8R^{2}(3+\Delta )-4R^{2}(3-\Delta )=$
$4R^{2}(6+2\Delta-3+\Delta)=$ $12R^{2}(1+\Delta )=$ $-48R^{2}\cos A\cos B\cos C\le 0$ $\Longrightarrow$ $\sum HL^{2}\le\frac{1}{4}\cdot\sum a^{2}\ .$

Remark. I posted long ago the following inequality : $\boxed{\ \sum IM^{2}\ge r(R+r)\ }\ ,$ where $I$ is the incenter of the triangle $ABC$ . Try solve it !

This post has been edited 1 time. Last edited by Virgil Nicula, Jan 24, 2012, 7:17 PM

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#3 h Feb 11, 2007, 8:25 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

A correction : $\sum IM^{2}\ge r(R+r)$

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#4 h Feb 11, 2007, 8:28 PM • 2 Y

Y by Adventure10, Mango247

I have wondered this for a long time: what does you a.s.o stand for? Is it something along the lines of "and symmetric others"?

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#5 h Feb 11, 2007, 9:52 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

$\left\{\begin{array}{c}4\cdot IM^{2}=2\left(IB^{2}+IC^{2}\right)-a^{2}\\\\ IB^{2}=\frac{ac(p-b)}{p}\\\\ IC^{2}=\frac{ab(p-c)}{p}\end{array}\right\|$ $\Longrightarrow$

$4p\cdot IM^{2}=2ac(p-b)+2ab(p-c)-pa^{2}=a[c(a+c-b)+b(a+b-c)-ap]=$

$a\left[(b-c)^{2}+a(b+c)-ap\right]=a(b-c)^{2}+a^{2}(p-a)$ $\Longrightarrow$

$\left\{\begin{array}{c}4p\cdot\sum IM^{2}=\sum \left[a(b-c)^{2}\right]+\sum a^{2}(p-a)\\\\ \sum a^{2}(p-a)=4S(R+r)\end{array}\right\|$ $\Longrightarrow$

$4p\cdot \sum IM^{2}=\sum \left[a(b-c)^{2}\right]+4pr(R+r)$ $\Longrightarrow$ $\boxed{\ \sum IM^{2}\ge r(R+r)\ }\ .$

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gemath

#6 h Feb 12, 2007, 10:37 AM • 2 Y

Y by Adventure10, Mango247

Forall $P$ on plane then $PA'^{2}+PB'^{2}+PC'^{2}\ge\frac{a^{2}+b^{2}+c^{2}}{12}\ge\frac{r(4R+r)}{3}\ge r(R+r)$ with $A',B',C'$ are midpoint of $BC,CA,AB$ resp
Thus the result of Virgil Nicula is true for all $P$ on plane! And more here we can show min of $\sum PA'^{2}$ with a fix triangle $ABC$ and really it is only simple result!

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tsemng

#7 h Feb 10, 2011, 8:54 AM • 2 Y

Y by Omeredip, Adventure10

Omid Hatami wrote:

In triangle $ABC$ , if $L,M,N$ are midpoints of $AB,AC,BC$ . And $H$ is orthogonal center of triangle $ABC$ , then prove that $LH^{2}+MH^{2}+NH^{2}\leq\frac14(AB^{2}+AC^{2}+BC^{2})$

i tried in complex co-ordinat and the result became
$2abc+ \sum (a^2b+b^2a) \ge 0$ is this true?

$a,b,c$ are co-ordinat of $A,B,C$ and $h=a+b+c$ $l=(a+b)/2$ $m=(a+c)/2$ $n=(b+c)/2$

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Potla

#8 h Feb 13, 2011, 8:55 PM • 2 Y

Y by Adventure10, Mango247

tsemng wrote:

Omid Hatami wrote:

In triangle $ABC$ , if $L,M,N$ are midpoints of $AB,AC,BC$ . And $H$ is orthogonal center of triangle $ABC$ , then prove that $LH^{2}+MH^{2}+NH^{2}\leq\frac14(AB^{2}+AC^{2}+BC^{2})$

i tried in complex co-ordinat and the result became
$2abc+ \sum (a^2b+b^2a) \ge 0$ is this true?

$a,b,c$ are co-ordinat of $A,B,C$ and $h=a+b+c$ $l=(a+b)/2$ $m=(a+c)/2$ $n=(b+c)/2$

Hmm, I am getting a different result.
Denote by $H$ , the origin; and give $A,B,C$ the affixes $u,v,w.$ So we see that $l=\frac{v+w}{2}, m=\frac{w+u}{2},n=\frac{u+v}{2}.$
Therefore $AB^2+BC^2+CA^2=(u-v)^2+(v-w)^2+(w-u)^2$ and
$LP^2+MP^2+NP^2=\frac 14\left[(w+u)^2+(u+v)^2+(v+w)^2\right].$
Also, $u+v+w=0$ ; so that
$\begin{aligned}\frac 14\left[AB^2+BC^2+CA^2\right]&=\frac 14\left[(u-v)^2+(v-w)^2+(w-u)^2\right]\\&=\frac 14\left[2(u+v+w)^2+(u-v)^2+(v-w)^2+(w-u)^2\right]\\&=\frac 14\left[2(u^2+v^2+w^2)+(u+v)^2+(v+w)^2+(w+u)^2\right]\\&\geq \frac 14\left[(u+v)^2+(v+w)^2+(w+u)^2\right];\end{aligned}$
And we are done. $\Box$

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#9 h Jan 24, 2012, 6:47 PM • 2 Y

Y by Adventure10, Mango247

And so what is the equality case ? Is it when the triangle is reduced to a point ?
Thanks in advance.

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JustN

#10 h Feb 6, 2012, 5:19 PM • 5 Y

Y by MaththematicND, CrazyMathlete, Adventure10, Mango247, and 1 other user

Only if $ABC$ is acute or right. Equality holds if it is right.
We use the fact that reflection of the orthogonal center with the respect to the midpoints of the triangle lay on the circumcircle.
Denote $P,Q$ intersection of $HL$ with circumcircle. ( $P$ is reflection of $H$ wrt $L$ )
Looking at power of the point $L$ wrt to the circumcircle and noticing that $LP=HL \leq LQ$ we get
$\frac{AB^2}{4}=AL*BL=LQ*LP \geq HL^2$
Applying this to two other sides we get the desired result.

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Sardor

#11 h Feb 2, 2015, 11:25 AM • 2 Y

Y by Adventure10, Mango247

From Leybnits's identity it's equivalent to $cosA \cdot cosB \cdot cosC \ge 0$ .It's true if and only if $ABC$ is non-obtuse!

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2 posts

#12 h May 12, 2015, 6:16 AM • 1 Y

Y by Adventure10

Virgil Nicula wrote:

Only if the triangle $ABC$ is acute or right !

Omid Hatami wrote:

In acute or right triangle $ABC$ , if $L,M,N$ are midpoints of the its sides and $H$ is its orthocenter, then show that $\boxed{\ LH^{2}+MH^{2}+NH^{2}\leq\frac14\cdot \left(a^{2}+b^{2}+c^{2}\right)\ }$

Proof. It is well-known that $HA=2R|\cos A|$ , $a=2R\sin A$ a.s.o. and $\Delta\equiv\sum \cos 2A=-1-4\prod\cos A\ .$ Therefore,
$4\sum HL^{2}-\sum a^{2}=$ $\sum\left(4\cdot HL^{2}-a^{2}\right)=$ $\sum\left[2\left(HB^{2}+HC^{2}\right)-2a^{2}\right]=$ $4\sum HA^{2}-2\sum a^{2}=$
$16R^{2}\sum \cos^{2}A-8R^{2}\sum \sin^{2}A=$ $8R^{2}\sum (1+\cos 2A)-4R^{2}\sum (1-\cos 2A)=$ $8R^{2}(3+\Delta )-4R^{2}(3-\Delta )=$
$4R^{2}(6+2\Delta-3+\Delta)=$ $12R^{2}(1+\Delta )=$ $-48R^{2}\cos A\cos B\cos C\le 0$ $\Longrightarrow$ $\sum HL^{2}\le\frac{1}{4}\cdot\sum a^{2}\ .$

Remark. I posted long ago the following inequality : $\boxed{\ \sum IM^{2}\ge r(R+r)\ }\ ,$ where $I$ is the incenter of the triangle $ABC$ . Try solve it !

Sorry,I don,t know why LH^2-a^2=2(BH^2+CH^2)-2a^2
Can you explain?

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#13 h May 12, 2015, 6:25 AM • 3 Y

Y by nicegeo, Adventure10, Mango247

$O$ the center of $\odot (ABC)$ and WLOG let $\odot (ABC)$ be unit circle vectors
By Euler line we get $H=a+b+c$

$LH^{2}+MH^{2}+NH^{2}\leq\frac14(AB^{2}+AC^{2}+BC^{2}) \Longleftrightarrow \sum_{cyc} (\frac{a+b+2c}{2})^2 \le \sum_{cyc} (\frac{a-b}{2})^2$

We use that $a^2=b^2=c^2=1$

$\Leftrightarrow -1 \ge ab+bc+ca$

By scalar product , $ab= cos(2C)$ ,

$\Leftrightarrow -1 \ge cos(2A)+cos(2B)+cos(2C)$

$\Leftrightarrow -1 \ge 2cos(A+B)cos(A-B)+2cos^2(A+B)-1$

$\Leftrightarrow 0 \ge +4cos(A+B)cosAcosB$

$\Leftrightarrow 0 \ge -4cos(C)cos(A)cos(B)$

This is true for acute or right triangle.

This post has been edited 1 time. Last edited by drmzjoseph, May 14, 2015, 5:34 PM

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