Highly recommended by the Problem Committee

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Highly recommended by the Problem Committee

G H J

Reveal topic tags

geometric inequality

L

G H BBookmark kLocked kLocked NReply

Source: IMO Shortlist 1993, Indonesia 1

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3647 posts

orl

#1 h Mar 25, 2006, 10:23 PM • 2 Y

Y by Adventure10, Mango247

The vertices $D,E,F$ of an equilateral triangle lie on the sides $BC,CA,AB$ respectively of a triangle $ABC.$ If $a,b,c$ are the respective lengths of these sides, and $S$ the area of $ABC,$ prove that

$DE \geq \frac{2 \cdot \sqrt{2} \cdot S}{\sqrt{a^2 + b^2 + c^2 + 4 \cdot \sqrt{3} \cdot S}}.$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

4 posts

#2 h Apr 7, 2006, 9:33 PM • 1 Y

Y by Adventure10

does anyone know how to solve this problem??
because i've tried a lot and i couldn't solve it...

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

2643 posts

yetti

#3 h Apr 14, 2006, 8:33 AM • 3 Y

Y by Illuzion, Adventure10, Mango247

The pedal triangle of the 1st isodynamic point Q (the intersection of the 3 Apollonius circles inside the triangle $\triangle ABC$ ) is equilateral. For example: Let D, E, F be the feet of normals from Q to BC, CA, AB. The A-Apollonius circle is a locus of points with the ratio of distances $\frac{AB}{AC}$ from the triangle vertices B, C and similarly for the B- and C-Apollonius circles. Hence $\frac{QB}{QC} = \frac{AB}{AC},$ $\frac{QC}{QA} = \frac{BC}{BA},$ $\frac{QA}{QB} = \frac{CA}{CB}.$ The quadrilateral AEQF is cyclic with the right angles at the vertices E, F, hence its circumradius is $R_A = \frac{AQ}{2},$ so that $EF = 2R_A \sin A = AQ \sin A$ and similarly $FD = BQ \sin B,$ $DE = CQ \sin C.$ Consequently,

$\frac{DE}{EF} = \frac{CQ \sin C}{AQ \sin A} = \frac{BC \sin C}{AB \sin A} = 1,\ \ DE = EF$

and similarly EF = FD. From the cyclic quadrilaterals AEQF, BFQD,

$\angle AQB = \angle AQF + \angle BQF = \angle AEF + \angle BDF =$

$= 180^\circ - \angle A - \angle AFE + 180^\circ - \angle B - \angle BFD =$

$= 180^\circ - (\angle A + \angle B) + 180^\circ - (\angle AFE + \angle BFD) = \angle C + 60^\circ$

and similarly, $\angle BQC = \angle A + 60^\circ,$ $\angle CQA = \angle B + 60^\circ.$ Let e = DE = EF = FD be the side length of the equilateral pedal triangle $\triangle DEF$ . The area S of the triangle $\triangle ABC$ with circumradius R is

$S = \frac 1 2 [AQ \cdot BQ\ \sin(C + 60^\circ) + BQ \cdot CQ\ \sin(A + 60^\circ) + CQ \cdot AQ\ \sin(B + 60^\circ)] =$

$= \frac{e^2}{2} [\frac{\sin(C + 60^\circ)}{\sin A \sin B} + \frac{\sin(A + 60^\circ)}{\sin B \sin C} + \frac{\sin(B + 60^\circ)}{\sin C \sin A} ] =$

$= \frac{4R^3e^2}{abc} [\sin A \sin(A + 60^\circ) + \sin B \sin(B + 60^\circ) + \sin C \sin(C + 60^\circ)] =$

$= \frac{R^2e^2}{S} [\frac 1 2 (\sin^2 A + \sin^2 B + \sin^2 C) + \frac{\sqrt 3}{2}(\sin A \cos A + \sin B \cos B + \sin C \cos C)] =$

$= \frac{e^2}{8S} [a^2 + b^2 + c^2 + \frac{4 \sqrt 3 R^2}{2}\ (\sin 2A + \sin 2B + \sin 2C)] =$

$= \frac{e^2}{8S} (a^2 + b^2 + c^2 + 4S \sqrt 3)$

$e = \frac{2S \sqrt 2}{\sqrt{a^2 + b^2 + c^2 + 4S \sqrt 3}}$

Thus the expression on the right side of the inequality in question is precisely the side length of the equilateral pedal triangle $\triangle DEF$ of the 1st isodynamic point Q. Any other equilateral triangle $\triangle D'E'F'$ inscribed in the triangle $\triangle ABC,$ so that $D' \in BC,\ E' \in CA,\ F' \in AB$ is obviously obtained from the equilateral pedal triangle $\triangle DEF$ by a spiral similarity with the center Q and similarity coefficient greater than 1, hence its side e' = D'E' is greater than the side e = DE.

Attachments:

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1693 posts

#4 h Aug 1, 2006, 11:54 AM • 2 Y

Y by Adventure10, Mango247

if $M$ is the fermat point of $ABC$ then the inequality reduces to :
$\sum |DE| \cdot |MC|\geq 2S$
this is kind of obvious because the product of the diagonals of a quadrilateral is $\geq$ twice its area...

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

2643 posts

yetti

#5 h Aug 4, 2006, 2:28 PM • 1 Y

Y by Adventure10

Not true in general, namely if the 1st Fermat point M is outside the triangle $\triangle ABC,$ say, if $\angle C > 120^\circ.$ Then the inequality is reduced to

$|EF| \cdot |MA|+|FD| \cdot |MB|-|DE| \cdot |MC|\geq 2S$

and the "obvious" argument fails.

The 1st isodynamic point Q is then also outside the triangle $\triangle ABC$ (inside its circumcircle) and the angle $\angle AQB,$ the convex one, is equal to

$\angle AQB = \angle AQF+\angle BQF = 180^\circ-\angle AEF+180^\circ-\angle BDF =$

$\angle A+\angle AFE+\angle B+\angle BFD = 180^\circ-\angle C+120^\circ = 360^\circ-(\angle C+60^\circ)$

so that the concave $\angle AQB = \angle C+60^\circ,$ as before. The area S of the triangle $\triangle ABC$ is still

$S = \frac{1}{2}\left[AQ \cdot BQ \sin(C+60^\circ)+BQ \cdot CQ \sin(A+60^\circ)+CQ \cdot AQ \sin(B+60^\circ)\right] =$

$... =\frac{e^{2}}{8S}\left(a^{2}+b^{2}+c^{2}+4S \sqrt 3\right)$

etc.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

4146 posts

#6 h Jun 27, 2009, 11:27 PM • 2 Y

Y by Adventure10, Mango247

It is known that among all the equilateral triangles circumscribed in $\triangle ABC,$ the triangle $\triangle XYZ$ homothetic to the outer Napoleon triangle of $\triangle ABC$ has the maximum area (side length). Hence, its side is twice the measure of the side $L$ of the outer Napoleon triangle.

$L = \sqrt {\frac {_1}{^6}(a^2 + b^2 + c^2) + \frac {2\sqrt {3}}{3}[\triangle ABC]}.$

For every circumcribed equilateral triangle $\Delta,$ we can associate an equilateral triangle $\Delta'$ homothetic to $\Delta$ and inscribed in $\triangle ABC.$ Thus, by Gergonne-Ann theorem, it follows that $[\triangle ABC]^2 = [\Delta][\Delta'].$ The area (side L') of $\Delta'$ will be minimum if the area (side) of $\Delta$ is maximum. In other words, $[\triangle ABC]^2 = \frac {_3}{^{16}}(L')^2(4L^2).$

Substituting the value of the side length $L$ of the Napoleon triangle yields:

$[\triangle ABC]^2 = \frac {_1}{^8}(L')^2(a^2 + b^2 + c^2 + 4\sqrt {3}[\triangle ABC])$

$\Longrightarrow L' = \frac {2\sqrt {2}[\triangle ABC]}{\sqrt {a^2 + b^2 + c^2 + 4\sqrt {3}[\triangle ABC]}}.$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

439 posts

#7 h Jun 28, 2009, 11:42 AM • 2 Y

Y by Adventure10, Mango247

Here is my solution:
Denote by $T$ the Toricelly point of triangle $ABC$ .Let $DE=EF=FD=x$
Appying Ptolemy's theorem
$S=[AEFT]+[CDTE]+[BDTF]$ ≤ $\frac{1}{2}(TA.EF+TC.DE+TB.FD)=\frac{x(TA+TB+TC)}{2}$
Hence $x$ ≥ $\frac{2S}{TA+TB+TC}=\frac{2S \sqrt 2}{\sqrt{a^2+b^2+c^2+4S \sqrt3}}$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

4146 posts

#8 h Jun 28, 2009, 5:48 PM • 2 Y

Y by Adventure10, Mango247

SUPERMAN2 wrote:

$S = [AEFT] + [CDTE] + [BDTF] \ldots$

This only works when the Fermat point lies inside ABC. See post #5

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

293 posts

#10 h Jan 29, 2021, 3:30 PM

Y by

Its well known that orthic triangle has least perimeter of all inscribed triangles. Thus $x>\frac{1}{3}\Sigma a cos\alpha$ . So if we show $\frac{1}{3}\Sigma a cos\alpha \ge \frac{2S \sqrt 2}{\sqrt{a^2 + b^2 + c^2 + 4S \sqrt 3}}$ Is this inequality true?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

82 posts

596066

#11 h Apr 21, 2021, 7:29 PM • 1 Y

Y by Mango247

My rough solution is exactly same as done in IMO compendium. Assuming that all the angles of the triangle $ABC$ are less than $120^\circ$ . Let the point $T$ be the Torricelli point of the triangle. So, $\angle(AT,EF)=\angle(BT,FD)=\angle(CT,DE)=\theta$ . So,
$2S= 2(S_{AETF}+S_{BFTD}+S_{CDTE})$
= $(AT+BT+CT)DE\sin\theta\le(AT+BT+CT)DE$

By cosine rule, $AT^2+AT.BT+BT^2=c^2$ ,
$BT^2+CT.BT+CT^2=a^2$ ,
$CT^2+CT.AT+AT^2=b^2$ .
And $3(AT.BT+BT.CT+CT.AT) =4\sqrt{3}(S_{ATB}+S_{BTC}+S_{CTA}) =4\sqrt{3}S$

Adding all, we get $2(AT+BT+CT)^2=a^2+b^2+c^2+4\sqrt{3}S$ .. This combined with the first inequality, gives the result.

Now let $\angle{C}\ge 120^{\circ}$ . Take T to be the point lying on the same side of $AB$ as $C$ such that $\angle BTC=\angle CTA= 60^{\circ}$
Now we have $2S\ge (AT + BT -CT)DE$ and $2(AT+BT-CT)^2=a^2+b^2+c^2+4\sqrt{3}S$ .

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a