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We are looking for angles x and y.

I have found the values of the following angles: BEA = 74, BDA = 64, ACD = 68, ECD = 112,

plus the relationship $x+y = 68$.

All other angles equations, from triangles or the sum of angles in the quadrilateral (360) end up in the same equation!

I have found through Geogebra that $x=18$ and $y=50$ but I can't figure out a second relationship to determine them geometrically!

Does anyone have any ideas?

Thank you!

Stevo
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Sal.Cognato
  • 1,527

4 Answers4

7

If you are looking for the proof by elementary geometry, please see Hiroshi Saito(斉藤浩)'s great work about generalized Langley's problem ( https://www.gensu.co.jp/saito/challenge/pdf/3circumcenter_d20180609.pdf ). He introduced the amazing skill named "3 circumcenter method" invented by Ms. aerile_re(pen name), and you can find the solution of this problem in the article (Q1).

Ichungchen
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  • Although it is impossible for me to follow this solution, it is indeed the geometrical solution to the problem! Thank you lchungchen! – Sal.Cognato Oct 27 '20 at 07:35
2

my geometric solution: http://www.davdata.nl/math/geopuzzle28.html

arc calculation

arc calculation(2)

David Dirkse
  • 21
0

A proof by Sine Law goes as follows:

Considering triangles $AED, BED, ABD, ABE$:

\begin{align} \frac {\sin y}{AD} &= \frac {\sin 48^\circ}{ED}\\ \frac {\sin x}{EB} &= \frac {\sin 38^\circ}{ED}\\ \frac {\sin 46^\circ}{AD} &= \frac {\sin 64^\circ}{BA}\\ \frac {\sin 22^\circ}{EB} &= \frac {\sin 74^\circ}{BA}\\ \end{align}

Equating $ED$ in the first two equations we have:

$$\frac {EB \sin 38^\circ}{\sin x} = \frac {AD \sin 48^\circ}{\sin y}$$

$EB$ and $AD$ can be expressed in terms of $BA$:

$$\frac {AB \sin 22^\circ \sin 38^\circ}{\sin 74^\circ \sin x} = \frac {AB \sin 46^\circ \sin48^\circ}{\sin 64^\circ \sin y}$$

Noting that $x = 68^\circ - y$,

$$\frac {\sin 22^\circ \sin 38^\circ}{\sin 74^\circ (\sin 68^\circ \cos y - \cos 68^\circ \sin y)} = \frac {\sin 22^\circ \sin 38^\circ}{\sin 74^\circ \sin (68^\circ - y)} = \frac {\sin 46^\circ \sin48^\circ}{\sin 64^\circ \sin y}$$

Rearranging:

$$(\sin 22^\circ \sin 38^\circ \sin 64^\circ + \sin 46^\circ \sin 48^\circ \sin 74^\circ \cos 68^\circ) \sin y = \sin 46^\circ \sin 48^\circ \sin 74^\circ \sin 68^\circ \cos y$$

Giving the expression for $\tan y$:

$$\frac {\sin 46^\circ \sin 48^\circ \sin 74^\circ \sin 68^\circ} {\sin 22^\circ \sin 38^\circ \sin 64^\circ + \sin 46^\circ \sin 48^\circ \sin 74^\circ \cos 68^\circ}$$

WolframAlpha says it is $50^\circ$. One could probably reduce the expression above, but right now I don't see how.

player3236
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-1

Sinx / Sin(x + 64) = Sin38 / Sin84 *Sin22 / Sin48

Multiply both denom and nom by 4Sin82

Sinx /Sin(x + 64) = 4Sin22Sin38Sin82 / (Cos6Sin48Sin82)

(4Sin22Sin38Sin82 = Sin66 REPLACEMENT)

Sinx / Sin(x + 64) = Cos24 / (4Sin48Cos6Sin82)

Sinx / Sin(x + 64) = 1 / (8Sin24Cos6Sin82)

(2Sin24*Cos6 = Sin54 REPLACRMENT)

Sinx / Sin(x + 64) = 1 / (4Sin54*Sin82)

(1/4Sin54 = Sin18 REPLACEMENT)

Sinx / Sin(x + 64) = Sin18 / Sin82

X = 18

Y = 50

Hakan Ersöz
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