Basicly, this problem is very easy . I think it would be more suitable for Beginer's Section. Here is the solution: Divide the square in 4 smaller squares, of area 1/4. By Dirichlet, we have in one of them 3 points. Now show that the maximum area is 1/8. Do this by drawing parallel lines to the sides of the square and move the points. Try yourself, now that you know the ideas...

In fact I've posted it at the Advanced Section Section accidentally,although it seemed very difficult to me. What is Dirichlet?

Dirichlet stands for the Dirichlet principle, also called Pigeonhole Principle. Roughly speaking, this principle says that if I have 15 rabbits in 14 cages, then there must be at least one cage with more than one rabbit. It's one of the fundamental principles of number theory and combinatorics. Darij

Well, in fact Dust you only proved that a triangle has area $ \leq \frac {1}{8}$, which is indeed easy. The problem asked for a strict inequality, thus there is some more work to do.... Pierre.

pbornsztein wrote: The problem asked for a strict inequality, thus there is some more work to do.... There are lots of cases!! Well, let's call A, B, C, D the four small squares, in clockwise order, and suppose all the triangles with vertices in 3 of our 9 points have area >= 1/8. If 3 points lie inside a square, then two of them must lie on the vertices of one of its sides, while the other one must lie on the opposite side. This is easy to prove by DusT's outlined argument. If 4 points lie inside a square, then they must coincide with its vertices. More than 4 points cannot lie inside the same square, or they'll generate a too small triangle. Now, suppose that exactly 4 points lie inside A. Then, the remaining 5 points can lie on just two sides of the big square: the one touching B and C, and the one touching C and D. By pigeonhole principle, 3 points must lie on the same side, generating a triangle with area 0. So, not more than 3 points can lie inside the same square. Suppose that exaclty 3 points lie inside A. Then, by the above argument both B and D have at least one point in common with A (meaning that the points lie on the common sides). Then, of the remaining 6 points, not more than 2 can lie inside B, and the same for D. Since not more than 3 of them can lie inside C, then exactly two must lie inside B or D. Suppose by symmetry it's B: then they must lie on the distant vertices wrt A, and one of them must be in common with C. A similar result holds for D, and so 4 points must lie in C: contradiction.

I known there were some cases, and I was too lazy to check them by myself. So, thanks for doing the job Mindflyer Pierre.

I think it worths mentioning : Show that given any m^2 points inside a square of side 1 we can always find 3 which form a triangle with area less than $\frac{1}{2(m-1)^2} $ .

I think it can be proven immediately because, sincethe points are IN the square, we can make a smaller square that includes the 9 points, and by Dirichlet again the area is less than 1/8 P.S mahanmath, you listen to linkin park?

Can someone please show me in more detail the argument that the maximum possible area of a triangle within one of the smaller squares is 1/8?

mahanmath wrote: I think it worth mentioning : Show that given any $m^2$ points inside a square of side $1$ we can always find $3$ which form a triangle with area less than $\dfrac{1}{2(m-1)^2} $ . See http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1030132&sid=882f23ec68fcc2399e09ec3ff54a1394#p1030132 for origin and solution. Notice the naive attempt to again use the pigeonhole principle fails for $m>3$. On the other hand, a slight refinement of the method in the link (printed in the RMC 2008) proves that for $m=3$, i.e $9$ points, one can find a triangle of area at most $\dfrac {1} {9}$.

I don't really understand mindflyer's solution. Could someone please explain it? Thanks.

cheeseyicecream wrote: Can someone please show me in more detail the argument that the maximum possible area of a triangle within one of the smaller squares is 1/8? Bump, I would appreciate one too.

[asy][asy] import graph; size(4cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; real xmin = -4.3, xmax = 11.92, ymin = -2.22, ymax = 6.3; pen zzttqq = rgb(0.6,0.2,0); pen ffqqtt = rgb(1,0,0.2); draw((-2,4)--(-2,-1)--(3,-1)--(3,4)--cycle, zzttqq); draw((-2,4)--(-2,-1), zzttqq); draw((-2,-1)--(3,-1), zzttqq); draw((3,-1)--(3,4), zzttqq); draw((3,4)--(-2,4), zzttqq); draw((-0.24,3.82)--(-1.46,1)); draw((-1.46,1)--(1.98,-0.44)); draw((1.98,-0.44)--(-0.24,3.82)); draw((-2,1)--(3,1), ffqqtt); draw((-0.24,3.82)--(-0.24,1)); draw((1.98,1)--(1.98,-0.44)); dot((-2,4),dotstyle); label("$A$", (-1.92,4.12), NE * labelscalefactor); dot((-2,-1),dotstyle); label("$B$", (-1.92,-0.88), NE * labelscalefactor); dot((3,-1),dotstyle); label("$C$", (3.08,-0.88), NE * labelscalefactor); dot((3,4),dotstyle); label("$D$", (3.08,4.12), NE * labelscalefactor); dot((-0.24,3.82),dotstyle); label("$E$", (0,3.64), NE * labelscalefactor); dot((-1.46,1),dotstyle); label("$F$", (-1.46,1.32), NW * labelscalefactor); dot((1.98,-0.44),dotstyle); label("$G$", (2.06,-0.32), NE * labelscalefactor); dot((-2,1),dotstyle); label("$H$", (-2.44,0.92), W * labelscalefactor); dot((3,1),dotstyle); label("$I$", (3.26,1), NE * labelscalefactor); dot((1.23,1),dotstyle); label("$J$", (1.3,1.12), NE * labelscalefactor); dot((-0.24,1),dotstyle); label("$K$", (-0.16,1.12), NE * labelscalefactor); dot((1.98,1),dotstyle); label("$L$", (2.06,1.12), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Suppose we have 3 arbitrary points $E, F, G$ inside (or on) the unit square. If all 3 points are on the vertices of the uniq square, then we know the area of the triangle in $\frac{1}{2}$. Otherwise, pick a point such that a horizontal or vertical line through it will cut the triangle into two triangles. For example, line $HI$ in the diagram cuts $\triangle EFG$ into $\triangle EFJ$ and $\triangle FGJ$. It is easy to see that: \[ [EFG] = [EFJ] + [FGJ] = \frac{1}{2} FJ (EK + LG) \le \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}. \]

It boils down to showing that given any 3 points inside a square with area $A$, with points being allowed on the boundary only in the pink part (see Fig.2 attatched--bottom one), then the area of the triangle formed by those points is strictly less than $\tfrac{A}{2}$. We can then apply this reasoning to the problem by splitting the given unit square into 4 congruent ones each with area $\tfrac{1}{4}$, then by the Pigeonhole principle one of these squares must contain 3 points. Now the problem states that points must lie within the unit square, so the points can other be on the pink part or inside the square (See Fig.1--top one), then by the above lemma the area of any triangle formed by 3 such points is strictly less than $\tfrac{A}{2}=(\tfrac{1}{4})/2=\tfrac{1}{8}$.

divide the square into 8 parts, and the 9 points means there are 2 in a part so it's at most area 1/8

nope, i disagree

I think that we don't have to consider cases which some points lie on a side of large square. because the problem states that points lie inside a square if not, then what about the case which all 9 points lie on 9 vertices of small squares when we divide into 2x2

Simply split the square into four smaller squares with area $\frac{1}{4}.$ By the Pigeonhole Principle, one of these squares must contain 3 points and the largest possible triangle in such a square has area $\frac{1}{8}.$ However, I don't get how to go from here to proving the strict inequality.

@above, do you mean each must contain 3 or more points?

twinbrian wrote: @above, do you mean each must contain 3 or more points? Yup.

Mathlete2017 wrote: Simply split the square into four smaller squares with area $\frac{1}{4}.$ By the Pigeonhole Principle, one of these squares must contain 3 or more points and the largest possible triangle in such a square has area $\frac{1}{8}.$ However, I don't get how to go from here to proving the strict inequality. Bump.

The points lie inside the square, and thus cannot be on the square. We can choose a smaller square where at least one of the points is on the square. Its area must be less than that of the square, so by the above argument, the area of the triangle is at most $\frac18$ of the area of the small square and is thus less than $\frac18$ the area of the large square. edit: actually I think this is wrong. They can be on the square.

It says strictly inside (supposed no 3 of them are collinear)

Mathlete2017 wrote: Simply split the square into four smaller squares with area $\frac{1}{4}.$ By the Pigeonhole Principle, one of these squares must contain 3 or more points and the largest possible triangle in such a square has area $\frac{1}{8}.$ However, I don't get how to go from here to proving the strict inequality. Bump.

A simple solution: Divide the square in $4$ right angled triangle (see the picture). We apply the Pigeonhole Principle. We can see that in on of the $4$ triangle there exist definitely $3$ points, let this triangle the $ABC$ triangle. But the area of the triangle $ABC$ it is exactly $1/8$

TuZo wrote: A simple solution: Divide the square in $4$ right angled triangle (see the picture). We apply the Pigeonhole Principle. We can see that in on of the $4$ triangle there exist definitely $3$ points, let this triangle the $ABC$ triangle. But the area of the triangle $ABC$ it is exactly $1/8$ If you divide a unit square into 4 congruent pieces, the area of each piece must be $\frac14$.

Mathlete2017 wrote: Mathlete2017 wrote: Simply split the square into four smaller squares with area $\frac{1}{4}.$ By the Pigeonhole Principle, one of these squares must contain 3 or more points and the largest possible triangle in such a square has area $\frac{1}{8}.$ However, I don't get how to go from here to proving the strict inequality. Bump.

in the case of largest possible triangle that is one of area 1/8 at least one of the vertices will lie on the exterior square (square of side 1) so that case will not be considered (inside means within the area enclosed by the sides and not on them) thus area will be less than 1/8

I'm kinda having problems with this? Sorry not the Pigeonhole master. I divided it into four congruent squares with area $\frac{1}{4}$ and so at least one must have $3$ points in it and the maximum area is $\frac{1}{8}$. However the problem says less, so what do I do? Thanks for the help.

Can I get some help? Thanks.

I solved it. Posting for storage.

Label the square $ABCD$, and let the midpoints of $AB,BC,CD,DA$ be $E,F,G,H$ respectively, and let the center of $ABCD$ be $O$. Then by Pigeonhole we may assume WLOG that some three of the points lie either in the interior of $AEOH$ or on $EO,HO$. So the area is less than or equal to $\frac{1}{8}$, but equality cannot occur because this implies that one of the points must be $A,E,H$ by Pigeonhole, contradiction.