# Geometry: Auxiliary Lines

Geometry: Auxiliary Lines

Would be very interesting to see a geometric solution.

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Almost the same as a question you asked after this one: math.stackexchange.com/questions/228700/… –  Joel Reyes Noche Nov 4 '12 at 11:07

Proof is almost the same as in Finding an unknown angle

Let's call central point $O$.

$\angle OBC = 180^\circ - (24^\circ+26^\circ) - (28^\circ+10^\circ) - X = 92^\circ-X$.

Using the sine formula, we have: $$\frac{OA}{\sin X} = \frac{OB}{\sin 24^\circ}, \quad \frac{OB}{\sin 10^\circ} = \frac{OC}{\sin (92^\circ-X)}, \quad \frac{OC}{\sin 26^\circ} = \frac{OA}{\sin 28^\circ},$$ hence $$OA = OB \cdot \frac{\sin X}{\sin 24^\circ} = OC \cdot \frac{\sin 10^\circ}{\sin(92^\circ-X)} \cdot \frac{\sin X}{\sin 24^\circ} = OA \cdot \frac{\sin 26^\circ}{\sin 28^\circ} \cdot \frac{\sin 10^\circ}{\sin(92^\circ-X)} \cdot \frac{\sin X}{\sin 24^\circ},$$ so, $$\sin 26^\circ \cdot \sin 10^\circ \cdot \sin X = \sin 28^\circ \cdot \sin (92^\circ-X) \cdot \sin 24^\circ. \qquad (1)$$

If $X \in \left[90^\circ, 92^\circ \right]$, then (1) has no solutions
(because $\sin X \ge \sin 88^\circ > \sin 28^\circ$, $\sin(92^\circ-X)\le\sin 2^\circ < \sin 10^\circ$ ).

If $X \in \left[0,90^\circ\right]$, then left side is ascending function, right side is descending function, so solution is unique (if exists).

We'll show that that unique solution of equation (1) is $X=70^\circ$. So, we need to prove equality $$\sin 26^\circ \cdot \sin 10^\circ \cdot \sin 70^\circ = \sin 28^\circ \cdot \sin 22^\circ \cdot \sin 24^\circ. \qquad (2)$$

Using formulas $$2\cdot \sin\alpha \cdot \sin\beta = \cos(\alpha-\beta) - \cos(\alpha+\beta),$$ $$2\cdot \sin\alpha \cdot \cos\beta = \sin(\alpha-\beta) + \sin(\alpha+\beta),$$ $$\sin(180^\circ-\varphi) = \sin(\varphi), \quad \cos(180^\circ-\varphi) = -\cos(\varphi),$$

we have $$\sin 26^\circ \cdot (\cos 60^\circ - \cos 80^\circ) =^{?} \sin 28^\circ \cdot (\cos 2^\circ - \cos 46^\circ),$$ $$-\sin 34^\circ + \sin 86^\circ + \sin 54^\circ - \sin 74^\circ =^{?} \sin 26^\circ + \sin 30^\circ + \sin 18^\circ - \sin 74^\circ,$$ $$-\sin 34^\circ + \sin 86^\circ + \sin 54^\circ - \sin 26^\circ - \sin 30^\circ - \sin 18^\circ =^{?} 0,$$ $$(-\sin 34^\circ + \sin 86^\circ - \sin 26^\circ) + (\sin 54^\circ - \sin 18^\circ -\frac{1}{2} )=^{?} 0,$$ $$(-\sin34^\circ + \sin 86^\circ + \sin 206^\circ) + \frac{1}{2} ( - \sin 18^\circ + \sin 54^\circ +\sin 126^\circ + \sin 198^\circ + \sin270^\circ )=^{?} 0,$$ it is equality, because left-hand side is equal to $$\mathrm{Im} \left( p(\omega_3^0 + \omega_3^1 + \omega_3^2) + \frac{q}{2}( \omega_5^0 + \omega_5^1 +\omega_5^2 + \omega_5^3 + \omega_5^4) \right)=$$ $$\mathrm{Im} \left( p \cdot \frac{\omega_3^3-1}{\omega_3-1} + \frac{q}{2}\cdot \frac{\omega_5^5-1}{\omega_5-1} \right)= \mathrm{Im} \left( p \cdot \frac{1-1}{\omega_3-1} + \frac{q}{2}\cdot \frac{1-1}{\omega_5-1} \right)= \mathrm{Im} \left( p \cdot 0 + \frac{q}{2} \cdot 0 \right)= 0,$$ where $p = \exp(-i\pi 17/90)$, $q = \exp(-i\pi /10)$, $w_{...}$ $-$ roots of unity:
$w_3 = \exp(i 2 \pi / 3) = \cos 2\pi/3 + i \sin 2\pi/3 = \cos 120^\circ + i \sin 120^\circ$,
$w_5 = \exp(i 2 \pi / 5) = \cos 2\pi/5 + i \sin 2\pi/5 = \cos 72^\circ + i \sin 72^\circ$.

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There is a triangle 26-28-126; the central angle is A+B+126=360, and the other triangles give $A+X+24=180; B+Y+10=180; X+Y+24+26+28+10=180$.

Solving the equations gives: $A=64+Y, B=170-Y, X=92-Y$.

So you need to do more to solve this, for example use the law of sines.

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Use the other equation $A+B+126=360$. You now have four equations in four unknowns. –  Joel Reyes Noche Nov 8 '12 at 3:11
@JoelReyesNoche That follows from the formulas for A and B. –  i. m. soloveichik Nov 8 '12 at 3:25
Oops, didn't see that. Thanks! –  Joel Reyes Noche Nov 8 '12 at 3:34

You have a basic problem - given three tuples of angles $a_1, b_1, c_1$; $a_2, b_2, c_2$; and $a_3,b_3$and $c_3$ - such that they are angles of 3 triangles (that is $a_i+b_i+c_i=180$; the triangles are $A_1B_1C_1$, $A_2B_2C_2$ and $A_3B_3C_3$, all determined up to scale) and such that the 3 vertices $C_1, C_2$ and $C_3$ of these triangles can be put together to fill the plane ($c_1+c_2+c_3=360$), when can these triangles be scaled so that the other 6 vertices match up in pairs to form a big triangle?

We can match up $B_1$ with $A_2$ by rescaling, and then match up B_2 with $A_3$. To then have $B_3$ match up with $A_1$ it is necessary and sufficient to have

$\sin A_1 \sin A_2 \sin A_3 = \sin B_1 \sin B_2 \sin B_3$

as you can see by repeatedly applying the theorem of sines to the triangles $A_iB_iC_i$ as in Oleg567's answer.

To get back to your problem, you can use this general fact to check that X=70 works. Namely, you form triangles 26-28-126, 70-24-86 and 10-22-148 and see that since $\sin 28 \sin 24 \sin 22= \sin 10 \sin 70 \sin 26$ the arrangement works.

Finally you know that the value of X that works is unique. This is easy to see: you start with a 50-38-92 triangle, draw the line dividing the 50 angle into 24 and 26, and draw the line dividing the 38 angle into 28 and 10, they determine an intersection point P, that point determines the angle X uniquely - the whole picture is canonical up to scale. Hence in the picture you have X=70 indeed.

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Well, first find D which is just basic arithmetic. Then, solve for every other unknown angle in terms of $X$ and then solve for $X$ accordingly. I am confused as to what building you are talking about.

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