Related-to-Q16140-What-if-the-three-lines-d-1-d-2-d-3-are-not-parallel-but-concurrent-

Question Number 16302 by mrW1 last updated on 20/Jun/17

Related to Q 16140

What if the three lines d_{1}, d_{2}, d_{3} are

not parallel, but concurrent ?

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/Jun/17

w h e r e i s^{'} b^{'} ?

Commented by mrW1 last updated on 20/Jun/17

Commented by mrW1 last updated on 20/Jun/17

positions of lines are given by α and β .

position of point A is given by a .

position of B and C is to determine .

side length of equilateral triangle

Δ ABC is to determine .

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/Jun/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/Jun/17

Answered by ajfour last updated on 21/Jun/17

s i d e l e n g t h o f e q u i l a t e r a l Δ A B C

i s g i v e n b y (w h e r e m_{1} = \tan α,

a n d m_{2} = \tan β)

s = \frac{2 a \sqrt{m_{1}^{2} m_{2}^{2} + m_{1}^{2} + m_{2}^{2} + m_{1} m_{2} - \sqrt{3} m_{1} m_{2} (m_{1} + m_{2})}}{\sqrt{3} (1 - m_{1} m_{2}) + (m_{1} + m_{2})} .

Commented by ajfour last updated on 20/Jun/17

Commented by ajfour last updated on 20/Jun/17

A i s t a k e n o n t h e r i g h t .

∠ O A B = θ, ∠ O A C = π / 3 - θ

\tan α = \frac{s \sin θ}{a - s \cos θ} = m_{1} \dots (i)

\tan β = \frac{s \sin (π / 3 - θ)}{a - s \cos (π / 3 - θ)} = m_{2}

\dots (i i)

f r o m (i) :

s \sin θ = {a m}_{1} - {s m}_{1} \cos θ

s = \frac{{a m}_{1}}{\sin θ + m_{1} \cos θ} \dots = (i i i)

f r o m (i i) :

s = \frac{{a m}_{2}}{\sin (π / 3 - θ) + m_{2} \cos (π / 3 - θ)}

\dots . (i v)

w e s h a l l f i n d s f r o m (i i i) w h e n

w e f i n d θ . A n d t o f i n d θ w e s h a l l

e q u a t e (i i i) a n d (i v) :

m_{1} [\sin (π / 3 - θ) + m_{2} \cos (π / 3 - θ)]

= m_{2} [\sin θ + m_{1} \cos θ]

m_{1} [\frac{\sqrt{3}}{2} \cos θ - \frac{\sin θ}{2}] + m_{1} m_{2} [\frac{\cos θ}{2} + \frac{\sqrt{3} \sin θ}{2}]

= m_{2} \sin θ + m_{1} m_{2} \cos θ

(\cos θ) [\frac{\sqrt{3} m_{1}}{2} - \frac{m_{1} m_{2}}{2}] =

(\sin θ) [\frac{m_{1}}{2} - \frac{m_{1} m_{2} \sqrt{3}}{2} + m_{2}]

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

F r o m (i i i)

s = \frac{{a m}_{1}}{\sin θ + m_{1} \cos θ}

= \frac{{a m}_{1} \sqrt{p^{2} + b^{2}}}{p + m_{1} b}

= \frac{{a m}_{1} \sqrt{p^{2} + b^{2}}}{m_{1} (\sqrt{3} - m_{2}) + m_{1} (m_{1} + 2 m_{2} - \sqrt{3} m_{1} m_{2})}

= \frac{a \sqrt{p^{2} + b^{2}}}{\sqrt{3} (1 - m_{1} m_{2}) + (m_{1} + m_{2})}

\sqrt{p^{2} + b^{2}} c o m e s o u t t o b e

= 2 \sqrt{(m_{1}^{2} m_{2}^{2} + m_{1}^{2} + m_{2}^{2} + m_{1} m_{2} - \sqrt{3} m_{1} m_{2} (m_{1} + m_{2})}

s = \frac{2 a \sqrt{m_{1}^{2} m_{2}^{2} + m_{1}^{2} + m_{2}^{2} + m_{1} m_{2} - \sqrt{3} m_{1} m_{2} (m_{1} + m_{2})}}{\sqrt{3} (1 - m_{1} m_{2}) + (m_{1} + m_{2})}

T o l o c a t e B a n d C w e n e e d o n l y

t o d r a w A B a n d A C a t a n g l e s

θ a n d (π / 3 - θ) w i t h A O r e s p e c t i v e l y

f r o m p o i n t A . T h e n w e j o i n B C .

θ = \tan^{- 1} (\frac{m_{1} (\sqrt{3} - m_{2})}{m_{1} + 2 m_{2} - \sqrt{3} m_{1} m_{2}})

T h e Δ A B C i s n o w i n p l a c e a n d

e q u i l a t e r a l .

Commented by ajfour last updated on 20/Jun/17

I s i t c o r r e c t S i r ?

I c h e c k e d i t o n l y f o r α = π / 6

a n d β = 0 . I t a n s w e r s w e l l;

e q u a l t o a / 2 (t h a t i s c o r r e c t i t h i n k) .

i f α = π / 4, β = π / 4 e v e n t h e n

i t a n s w e r s w e l l . s = a (\sqrt{3} - 1)

= \frac{2 a}{\sqrt{3} + 1} = a \sqrt{4 - 2 \sqrt{3}}

A n d θ = π / 6 . (k i n d l y r e f e r t o t h e

d i a g r a m o f m y s o l u t i o n w h e n

c h e c k i n g t h e s e) .

Commented by mrW1 last updated on 20/Jun/17

I have no own calculation, so I can

not compare .

Your calculation seems to be correct .

I^{'} ll check the result later .

Generally there are two such triangles,

one to left side and one to right side

of point A .

Commented by mrW1 last updated on 21/Jun/17

Following picture shows how to

draw the requested triangles only

with ruler and compass .

Commented by mrW1 last updated on 20/Jun/17

Commented by ajfour last updated on 20/Jun/17

t b a n k y o u S i r .

Commented by mrW1 last updated on 21/Jun/17

How does your formula change for

the triangle on the richt side of point

A ?

Commented by ajfour last updated on 21/Jun/17

s = \frac{2 a \sqrt{m_{1}^{2} m_{2}^{2} + m_{1}^{2} + m_{2}^{2} + m_{1} m_{2} + \sqrt{3} m_{1} m_{2} (m_{1} + m_{2})}}{\sqrt{3} (1 - m_{1} m_{2}) - (m_{1} + m_{2})}

t h i s i s i t S i r, w h e n △ i s o n t h e

r i g h t s i d e o f p o i n t A .

i u n d e r s t a n d, t h e n α \to π - α

a n d β \to π - β, f o r t h e o t h e r c a s e .

Commented by mrW1 last updated on 21/Jun/17

absolutely smart work!

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