Let $ABCDEF$ be a convex hexagon of area $1$, whose opposite sides are parallel. The lines $AB$, $CD$ and $EF$ meet in pairs to determine the vertices of a triangle. Similarly, the lines $BC$, $DE$ and $FA$ meet in pairs to determine the vertices of another triangle. Show that the area of at least one of these two triangles is at least $3/2$.
Problem
Source: Balkan Mathematical Olympiad 2011. Problem 4.
Tags: inequalities, geometry, LaTeX, ratio, trigonometry, geometry proposed
jgnr
08.05.2011 10:32
This looks too simple for a problem 4 in BMO, please check my solution.
Suppose the opposite. Let (AB,CD),(CD,EF),(EF,AB), (DE,FA),(FA,BC),(BC,DE) meet at P,Q,R,S,T,U. Note that triangles PQR and STU are similar, assume that PQR is the smaller one. Let SE/SU=SF/ST=p, UD/US=UC/UT=q, TA/TS=TB/TU=r. If the area of STU is L, then we get that the area of ABCDEF is L(1-p^2-q^2-r^2)=1. Since we assume L<3/2, then p^2+q^2+r^2<1/3. Note that DE/PR >= (1-p-q) etc. Analogously as above, we get (1-p-q)^2+(1-q-r)^2+1(1-r-p)^2<1/3.
Let a=p+q, b=q+r, c=r+p. Then sum a^2 <= 4(p^2+q^2+r^2) < 4/3, and sum (1-a)^2 < 1/3. By CS, a+b+c <= sqrt(3 sum a^2). Let sqrt(3 sum a^2)=x.
1/3 > sum (1-2a+a^2) > 3-2x+x^2/3
From here we get 2<x<4, so sum a^2 > 4/3, a contradiction.
PS: sorry, I can't use latex now.
mavropnevma
08.05.2011 10:52
Quite correct; this approach (ratios of the segments, and ratios of areas expressed in terms of those) is quickly leading to the answer. The official solution works on showing that if we "fold" the six small triangles (created around the hexagon) towards the hexagon, they will cover it, so one of the two sets of three triangles belonging to each of the two large triangles will have an area of at least $1/2$ that of the hexagon.
salazar
08.05.2011 14:23
Does anyone know anything about the results? Thank you!
andreass
08.05.2011 18:17
mavropnevma wrote: Quite correct; this approach (ratios of the segments, and ratios of areas expressed in terms of those) is quickly leading to the answer. The official solution works on showing that if we "fold" the six small triangles (created around the hexagon) towards the hexagon, they will cover it, so one of the two sets of three triangles belonging to each of the two large triangles will have an area of at least $1/2$ that of the hexagon. Could you please post here the official solution?
Malik
11.05.2011 11:29
Here is my solution to this problem: First of all, notice that all the triangles in this figure are similar. Let $\Delta$ be the triangle with maximal area $K$, $\Delta'$ be the other big triangle with area $k'^2K$ where $k'\leq1$ is the ratio of similarity with respect to $\Delta$ and $k_1,k_2,\ldots,k_6$ are the ratio of similarity of the small triangles with respect to $\Delta$. The small triangles are ordered such that $\Delta_1,\Delta_2,\Delta_3$ are adjacent to one side of the triangle $\Delta$. We have the relations (the sum of sides of 3 small triangles=the side of a big triangle): $k_1+k_2+k_3=k_3+k_4+k_5=k_5+k_6+k_1=1$ and $k_2+k_3+k_4=k_4+k_5+k_6=k_6+k_1+k_2=k'$. and for the areas: $K+k'^2K=(k_1^2+\cdots+k_6^2)K+2$ which gives $K=\frac{2}{1+k'^2-(k_1^2+\cdots+k_6^2)}$. We want to prove that $K\geq 3/2$ which is equivalent to $3\sum k_i^2\geq3k'^2-1$. But: \[\begin{array}{lcl} 3\sum k_i^2&=&(k_1^2+k_2^2+k_3^2)+(k_2^2+k_3^2+k_4^2)+\cdots+(k_6^2+k_1^2+k_2^2)\\ &\geq&\frac{1}{3}(k_1+k_2+k_3)^2+\frac{1}{3}(k_2+k_3+k_4)^2+\cdots+\frac{1}{3}(k_6+k_1+k_2)^2\\ &=&1+k'^2\geq3k'^2-1 \end{array}\] since $k'\leq1$ and we are done. Clearly, this is easy for a problem 4.
dyd
16.05.2011 08:22
Isn't it this?
mavropnevma
16.05.2011 09:30
Sure enough, your Hexagram topic is fully equivalent to this problem. It would be interesting if you could reveal from what source you got the problem of your initial post? this in no way points to plagiarism re the BMO problem, since same mathematical truths may be (and are) re-discovered.
dyd
16.05.2011 23:13
To be honest, I was just playing with geogebra and discovered the fact. I'm also pretty sure that the proposer in no way create any kind of plagiarism by propose this problem, but I think the problem committee should be more thorough in selecting the problems. This forum is - in my opinion - the best source for training problems, so the last problem of this year's BMO would likely be a 'Yeah! I've seen this problem - on Mathlinks!'.
SKhan
22.06.2011 19:03
Let $EF$ and $AB$ meet at $P_1$, $AB$ and $CD$ meet at $Q_1$ and $CD$ and $EF$ meet at $R_1$.
Let $BC$ and $DE$ meet at $P_2$, $DE$ and $FA$ meet at $Q_2$ and $FA$ and $BC$ meet at $R_2$.
Let $\angle Q_1P_1R_1=\alpha$, $\angle R_1Q_1P_1=\beta$ and $\angle P_1R_1Q_1=\gamma$.
Let $p=\sin\alpha$, $q=\sin\beta$ and $r=\sin\gamma$
W.L.O.G. we can assume that $|FA|\leq |CD|$
Let $|FA|=pa$, $|BC|=qb$, $|DE|=rc$, $|CD|=px$, $|EF|=qy$ and $|AB|=rz$
Simple angle chasing shows that $\triangle P_1AF$, $\triangle BAR_2$, $\triangle BQ_1C$, $\triangle P_2DC$, $\triangle EDR_1$, $\triangle EQ_2F$, $\triangle P_1Q_1R_1$ and $\triangle P_2Q_2R_2$ are similar.
Sine rule on $\triangle P_1AF$: $\frac{|FP_1|}{q}=\frac{|P_1A|}{r}=\frac{|FA|}{p}=a$ so $|FP_1|=qa$ and $|P_1A|=ra$
Similarly, $|BQ_1|=rb$, $|Q_1C|=pb$, $|DR_1|=pc$, $|R_1E|=qc$, $|CP_2|=qx$, $|P_2D|=rx$, $|EQ_2|=ry$, $|Q_2F|=py$, $|AR_2|=pz$ and $|R_2B|=qz$
So $|Q_1R_1|=p(x+b+c)$, $|R_1P_1|=q(a+y+c)$, $|P_1Q_1|=r(a+b+z)$, $|Q_2R_2|=p(a+y+z)$, $|R_2P_2|=q(x+b+z)$ and $|P_2Q_2|=r(x+y+c)$
Sine rule on $\triangle P_1Q_1R_1$: $\frac{|Q_1R_1|}{p}=\frac{|R_1P_1|}{q}=\frac{|P_1Q_1|}{r} $ so $x+b+c=a+y+c=a+b+z$ so $x-a=y-b=z-c$
Let $d=x-a=y-b=z-c$
Then $|Q_1R_1|=p(a+b+c+d)$, $|R_1P_1|=q(a+b+c+d)$ and $|P_1Q_1|=r(a+b+c+d)$.
Also $|Q_2R_2|=p(a+b+c+2d)$, $|R_2P_2|=q(a+b+c+2d)$ and $|P_2Q_2|=r(a+b+c+2d)$.
Also $|FA|\leq |CD|$ so $pa\leq px$ so $d=x-a\geq 0$
Area of $ABCDEF=$ Area of $P_1Q_1R_1 -$ Area of $P_1AF -$ Area of $CQ_1B -$ Area of $EDR_1$
So $1=\frac{pqr(a+b+c+d)^2}{2} - \frac{pqra^2}{2} - \frac{pqrb^2}{2} - \frac{pqrc^2}{2}$
So $d^2 + 2d(a+b+c)+2ab+2bc+2ca=\frac{2}{pqr}$
So $\frac{3}{2}d^2+3d(a+b+c)+(ab+bc+ca)+(2ab+2bc+2ca)=\frac{3}{pqr}$
Now $(a-b)^2+(b-c)^2+(c-a)^2\geq 0\implies a^2+b^2+c^2\geq ab+bc+ca$
So $\frac{3}{2}d^2+3d(a+b+c)+(a^2+b^2+c^2)+(2ab+2bc+2ca)\geq \frac{3}{pqr}$
$d\geq 0$ so $4d^2+4d(a+b+c)+(a^2+b^2+c^2)+(2ab+2bc+2ca)\geq\frac{3}{pqr}$
So $(a+b+c+2d)^2\geq\frac{3}{pqr}$
So $\frac{pqr(a+b+c+2d)^2}{2}\geq \frac{3}{2}$
So area of $\triangle P_2Q_2R_2=\frac{pqr(a+b+c+2d)^2}{2}\geq \frac{3}{2}$
Thus the area of at least one of those two triangles is at least $\frac{3}{2}$.
Nari_Tom
23.01.2025 08:39
Suppose the opposite. Let the points \( P, Q, R, S, T, U \) be the intersections of the following lines: \[ (AB, CD), (CD, EF), (EF, AB), (DE, FA), (FA, BC), (BC, DE). \]Note that triangles \( PQR \) and \( STU \) are similar, with \( PQR \) being the smaller one. Let \[ \frac{SE}{SU} = p, \quad \frac{SF}{ST} = p, \quad \frac{UD}{US} = q, \quad \frac{UC}{UT} = q, \quad \frac{TA}{TS} = r, \quad \frac{TB}{TU} = r. \]If the area of triangle \( STU \) is \( L \), then we have the relation: \[ \text{Area of } ABCDEF = L(1 - p^2 - q^2 - r^2) = 1. \]Since we assume \( L < \frac{3}{2} \), it follows that: \[ p^2 + q^2 + r^2 < \frac{1}{3}. \]Note that: \[ \frac{DE}{PR} \geq (1 - p - q), \quad \text{etc.} \]Analogously, we get the inequality: \[ (1 - p - q)^2 + (1 - q - r)^2 + (1 - r - p)^2 < \frac{1}{3}. \] Let: \[ a = p + q, \quad b = q + r, \quad c = r + p. \]Then: \[ \sum a^2 \leq 4(p^2 + q^2 + r^2) < \frac{4}{3}, \]and: \[ \sum (1 - a)^2 < \frac{1}{3}. \] By the Cauchy-Schwarz inequality: \[ a + b + c \leq \sqrt{3 \sum a^2}. \]Let: \[ \sqrt{3 \sum a^2} = x. \] Thus: \[ \frac{1}{3} > \sum (1 - 2a + a^2) > 3 - 2x + \frac{x^2}{3}. \] From this, we get: \[ 2 < x < 4, \]which leads to the contradiction: \[ \sum a^2 > \frac{4}{3}. \]solution of: jgnr in LATEX
Nari_Tom
23.01.2025 09:39
So here is my motivation. Let $X, Y, Z, P, Q, R$ be the intersection of the sides $(AB,EF)$, $(AB,CD)$, $(DC,EF)$, $(AF,ED)$, $(AF,BC)$, $(BC,ED)$, respectively. Let $S$ be the area of the hexagon. Then we need to prove that at least one of the triangles $XYZ$ and $PQR$ have at least $3/2*S$ area. Since these triangles are similar, we can assume that one of them is bigger. Let's say $XYZ$. Then we can enlarge the $PQR$ until it becomes equal to &XYZ& by enlarging some of the sides. But this operation enlarges the $S$, but bigger triangle's area didn't changed, so from here we can deduce that in the worst case when these two triangles are equal. So it suffices to solve problem when $PQR$ and $XYZ$ are equal. (So that's trivial i guess)
Nari_Tom
23.01.2025 09:39
Another solution using stronger fact. Let $X, Y, Z, P, Q, R$ be the intersection of the sides $(AB,EF)$, $(AB,CD)$, $(DC,EF)$, $(AF,ED)$, $(AF,BC)$, $(BC,ED)$, respectively. Let $S$ be the area of the hexagon. And $x,y,z,p,q,r$ be the areas of $XAF, YBC, EDZ, PEF, QAB, RCD$. Then single claim finishes our problem. Claim: $q+y+r\geq S(ABCD)$. Proof: Let $ AB=m*BY $, $ CD=n*CY $ . Then it's suffices to show that: $ 1+m^2+n^2\geq mn+m+n $, which is trivial. So using this claim: $q+y+r\geq S(ABCD)$ and $x+p+z\geq S(AFED)$ which implies that $x+y+z+p+q+r\geq S(ABCDEF)=S$. So one of the sums $x+y+z$ and $p+q+r$ is at least $S/2$, which finishes the problem.