Question-16140

Question Number 16140 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 18/Jun/17

Answered by ajfour last updated on 18/Jun/17

{A r e a}_{Δ A B C} = \frac{1}{\sqrt{3}} (a^{2} + a b + b^{2})

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 18/Jun/17

t h a n k y o u s o m u c h m r A j f o u r . n i c e

m e t h o d . i l o v e i t .

c a n w e u s e t h i s s o l u t i o n f o r Q . 15969 ?

Commented by ajfour last updated on 18/Jun/17

Commented by ajfour last updated on 18/Jun/17

s \sin θ = a

s \sin (\frac{π}{3} - θ) = b

s (\frac{\sqrt{3}}{2} \cos θ - \frac{1}{2} \sin θ) = b

\sqrt{3} s \cos θ - s \sin θ = 2 b

\sqrt{3} \sqrt{s^{2} - a^{2}} = a + 2 b

3 (s^{2} - a^{2}) = {(a + 2 b)}^{2}

s^{2} = a^{2} + \frac{{(a + 2 b)}^{2}}{3}

A = \frac{\sqrt{3}}{4} [a^{2} + \frac{{(a + 2 b)}^{2}}{3}]

= \frac{1}{4 \sqrt{3}} (4 a^{2} + 4 b^{2} + 4 a b)

= \frac{1}{\sqrt{3}} (a^{2} + a b + b^{2}) .

Answered by mrW1 last updated on 18/Jun/17

When we fix the first vertex of an equilateral

on one line, {let}^{'} s say on line 2, and lay

the second vertex variably on an other line, {let}^{'} s

say on line 1, the track of the third

convex is a straight line . The inter -

section point of this line with the third

line is the second convex of the equi -

lateral triangle which we are searching .

Thus we can “ {draw}^{″} the requested equi -

lateral triangle in following way .

We fix the pint B on line 2 .

On line 1 we select at least two points

P_{1} and P_{2} . With {BP}_{1} and {BP}_{2} as base

we construct in the same direction

equilateral triangles {BP}_{1} Q_{1} and {BP}_{2} Q_{2} .

We connect Q_{1} and Q_{2} . This new line

intersects with line 3 at point C .

BC is then the one side of the requested

equilateral triangle . The point A on

line 1 can be then easily determined .

Commented by mrW1 last updated on 18/Jun/17

Commented by mrW1 last updated on 18/Jun/17

with d = length of sides

A = area of triangle

When we choose P_{1} and P_{2} so that

Q 1 lies on line 1 and Q_{2} lies on line 2,

we will see :

d^{2} = a^{2} + {(\frac{2}{\sqrt{3}} b + \frac{a}{\sqrt{3}})}^{2}

d^{2} = a^{2} + \frac{1}{3} {(2 b + a)}^{2}

= \frac{4}{3} (a^{2} + b^{2} + ab)

d = \frac{2}{\sqrt{3}} \sqrt{a^{2} + b^{2} + ab}

A = \frac{\sqrt{3}}{4} d^{2} = \frac{\sqrt{3}}{4} \times \frac{4}{3} (a^{2} + b^{2} + ab)

A = \frac{1}{\sqrt{3}} (a^{2} + b^{2} + ab)

Commented by RasheedSoomro last updated on 18/Jun/17

MrW 1, Mr ajfour & Mr Abu Behi .

What a depth in geometry you have!!

Commented by mrW1 last updated on 18/Jun/17

Thank you sir! Geometry has always

been my favorite subject .

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 18/Jun/17

t h a n k y o u s o m u c h d e a r m r W 1 . i a l w y e s

e n j o y y o u r a n s w e r s a n d c o m m e n e t s .

g o d b l e e s y o u m y m a s t e r .

d o y o u r e a d m y c o m m e n e t o n m r {A j f o u r}^{'} s

a n s w e r a b o u t u s i n g t h i s s o l u t i o n f o r

Q . 15969 ?

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 18/Jun/17

t h a n k y o u m r R a s h e e d . m y n a m e i s : a m i r .

n o o n e c a l l e d m e a s : A b u b e h i y e t . (h a h a h a)

Commented by mrW1 last updated on 18/Jun/17

To abu behi :

I have read you comment to Mr .

{ajfour}^{'} s solution .

In fact the three - line - problem is

similar to the three - circle - problem .

But the later is much more complicated .

For my solutions I used the same way

based on knowledge gained from the

graphics .

Now we have for both problems the

graphical and analytical solutions

(formula) . I have also added my

formula for Q 15969 . Please have a

look at it .

Or do you see anything which is in

your opinion still not solved yet ?

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 18/Jun/17

d e a r m r W 1 . t h a n k y o u f o r y o u r a t t e n t i o n .

n o, t h e r e i s n o d e f f e r e n c e b e t w e e n s o l u t i o n s

a n d e v r y t h i n g h a v e b e e n s o l v e d s o

n i c e .

b u t i s a i d t h a t w o r d s o n l y s u c h a s a n

i d e a a n d n o m o r e .

Commented by mrW1 last updated on 18/Jun/17

Thank you for all these interesting

geometry questions .

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