Question Number 14940 by mrW1 last updated on 05/Jun/17

For those who are interested in Geometry: A triangle has an area of 1 unit. Each of its sides is divided into 4 equal parts through 3 points. The first and the last point of each side will be connected with each other to form 2 inscribed triangles and these 2 triangles form a hexagon. Find the area of the hexagon. What is the result, if each side is equally divided into 5 parts, or generally into n parts?

Commented bymrW1 last updated on 05/Jun/17

Commented byb.e.h.i.8.3.4.1.7@gmail.com last updated on 06/Jun/17

Commented bymrW1 last updated on 06/Jun/17

you got X=((n(n−1)−3)/(n(n−1))) and X=(1/2) by n=3 but the right answer is X=2/9 by n=3

Commented bymrW1 last updated on 06/Jun/17

error in your working (I think): 2(S−X)≠((6S)/(n(n−1)))⇒X≠S−((3S)/(n(n−1)))

Commented byajfour last updated on 06/Jun/17

could it be =((49)/(160)) ?

Commented bymrW1 last updated on 06/Jun/17

((49)/(160)) for n=4 is correct, congratulation!

Commented byajfour last updated on 06/Jun/17

thank you Sir, this question of yours let me learn triangular coordinates all by myself..

Commented bymrW1 last updated on 06/Jun/17

I devised this question. That you got the right answer in such a short time shows that you have learnt quite a lot.

Commented bymrW1 last updated on 06/Jun/17

This question can be solved with basic geometry knowledge. It is not necessary to use trigonometry or analytic geometry.

Commented bymrW1 last updated on 06/Jun/17

To Behi′s father: you are very close at the final solution. I′m sure you′ll be able to get the general solution for case n.

Commented byb.e.h.i.8.3.4.1.7@gmail.com last updated on 07/Jun/17

S_1 =(1/2).(b/n).((n−1)/n).c.sinA=((n−1)/n^2 ).S S_1 ^′ =(1/2).((b′)/3).((c′)/3).sinA′=((S′)/9) S′=S−3×((n−1)/n^2 )S=((n^2 −3n+3)/n^2 )S S′_1 =((S′)/9)=((n^2 −3n+3)/(9n^2 )) S_(hexagon) =S′−3S′_1 =(((n^2 −3n+3)/n^2 )−3×((n^2 −3n+3)/(9n^2 )))S= =((2(n^2 −3n+3))/(3n^2 )).S for:n=3⇒S_h =((2×3)/(3×9)).S=(2/9).S

Commented bymrW1 last updated on 07/Jun/17

please check again: for n=4, the right answer should be S_h =((49)/(160))S (as Mr. ajfour evaluated) I think there is an error here: S′_1 ≠((S′)/9) (this is only true for the case n=3)

Answered by ajfour last updated on 07/Jun/17

((Area_(hexagon) )/(Area_△ )) =((2(n^2 −3n+3)^2 )/(n^3 (2n−3))) .

Commented bymrW1 last updated on 07/Jun/17

your answer is correct and you have used a different way as I originally expected. you applied analytic geometry. this is nice and easy, because you don′t need any tricks. but as I said one can also get the result without analytic geometry or trigonometry.

Commented bymrW1 last updated on 07/Jun/17

x=0.222 for n=3 x=0.306 for n=4 x→1 when n→∞ the hexagon gets bigger and bigger with increasing n

Answered by mrW1 last updated on 08/Jun/17

Here my solution without usage of trigonometry or analytic geometry. The most important rule which I use is shown in the next diagram. Area of ΔABC=(1/2)ah_A =(1/2) a b sin C Area of Δ123=(1/2)αah_1 =(1/2) αa βb sin C ⇒(A_(Δ123) /A_(ΔABC) )=αβ

Commented bymrW1 last updated on 08/Jun/17

Commented bymrW1 last updated on 08/Jun/17

Commented bymrW1 last updated on 08/Jun/17

Each side of triangle ΔILG is equally divided into n parts. n≥3. Let a=((LG)/n)=length of each part of LG Let us say the area of triangle ΔILG is S=1. According to the rule above we have A_(ΔACG) =((n−1)/n)×(1/n)×S=((n−1)/n^2 ) =A_(ΔCHI) =A_(ΔNAL) =A_(ΔBML) =A_(ΔMHI) =A_(ΔHBG) ⇒A_(ΔMHB) =A_(ΔAHG) =S−3×A_(ΔACG) =1−3×((n−1)/n^2 )=((n^2 −3n+3)/n^2 ) To get the area of the hexagon we now only need to get the area of the ♮redε as well as the ♮blackε small triangles.

Commented bymrW1 last updated on 08/Jun/17

Using the rule above we have A_(ΔACB) =((n−2)/n)×(1/n)×S=((n−2)/n^2 ) To get the area of A_(ΔDEB) we need to know how the line segment AC is divided by points D and E.

Commented bymrW1 last updated on 08/Jun/17

Commented bymrW1 last updated on 08/Jun/17

We introduce some auxiliary lines: CK parallel to IF CJ parallel to HB ΔCJG is similar to ΔHBG ⇒((JG)/(BG))=((CG)/(HG)) ⇒JG=((CG)/(HG))×BG=(1/(n−1))×BG=(a/(n−1)) ΔKCG is similar to ΔFIG ⇒((KG)/(FG))=((CG)/(IG)) ⇒KG=((CG)/(IG))×FG=(1/n)×((LG)/2)=(1/n)×((na)/2)=(a/2) ΔADF is similar to ΔACK ⇒((AD)/(AF))=((AC)/(AK)) ⇒AD=((AF)/(AK))×AC=((LF−LA)/(AG−KG))×AC =((((na)/2)−a)/(na−a−(a/2)))×AC=((n−2)/(2n−3))×AC ΔAEB is similar to ΔACJ ⇒((AE)/(AC))=((AB)/(AJ)) ⇒AE=((AB)/(AJ))×AC=((AB)/(AG−JG))×AC =((na−2a)/(na−a−(a/(n−1))))×AC=((n−1)/n)×AC DE=AE−AD=((n−1)/n)×AC−((n−2)/(2n−3))×AC =(((n−1)/n)−((n−2)/(2n−3)))×AC =(((n−1)(2n−3)−n(n−2))/(n(2n−3)))×AC =((n^2 −3n+3)/(n(2n−3)))×AC Using our rule abov we get A_(ΔDEB) =((DE)/(AC))×A_(ΔACB) =((n^2 −3n+3)/(n(2n−3)))×((n−2)/n^2 )=(((n^2 −3n+3)(n−2))/(n^3 (2n−3))) Area of hexagon: A_H =A_(ΔMHB) −3×A_(ΔDEB) A_H =((n^2 −3n+3)/n^2 )−3×(((n^2 −3n+3)(n−2))/(n^3 (2n−3))) =((n^2 −3n+3)/n^2 )[1−((3(n−2))/(n(2n−3)))] =((n^2 −3n+3)/n^2 )[((n(2n−3)−3(n−2))/(n(2n−3)))] =((n^2 −3n+3)/n^2 )[((2n^2 −6n+6)/(n(2n−3)))] =((2(n^2 −3n+3)^2 )/(n^3 (2n−3)))

Commented byb.e.h.i.8.3.4.1.7@gmail.com last updated on 08/Jun/17

Great job done mrW1.fantastic. I love this proof very much.

Commented byb.e.h.i.8.3.4.1.7@gmail.com last updated on 08/Jun/17

after this ,i am waiting for mr Ajfour′s proof too.