Wikipedia:Reference desk/Archives/Mathematics/2018 May 6

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May 6

Set of trigonometric ratios for a triangle

If the following trigonometric ratios associated to a triangle, formed by dividing the lengths of the sides to the sines of opposites angles to the sides, are constant and equal between them, how are the ratios involving lengths of the sides and other trigonometric functions of opposite angles like cosines, tangents, cotangents? Thanks.--82.137.13.3 (talk) 13:37, 6 May 2018 (UTC)

See law_of_cosines, law_of_tangents and law_of_cotangents. Ruslik_Zero 20:28, 6 May 2018 (UTC)

Interesting the law of cotangents. It may suggest that cotangents ratios are also equal.--82.137.11.1 (talk) 10:51, 7 May 2018 (UTC)

But does the equality of the sine ratios ${\frac {a}{\sin A}}\,=\,{\frac {b}{\sin B}}\,=\,{\frac {c}{\sin C}}\$ imply the equality of cosines ratios

${\frac {a}{\cos A}}\,=\,{\frac {b}{\cos B}}\,=\,{\frac {c}{\cos C}}\$?--82.137.13.6 (talk) 11:00, 7 May 2018 (UTC)

No, because $\sin \theta =\cos {(\pi /2-\theta )}$. Iffy★Chat -- 11:35, 7 May 2018 (UTC)

The same for the equality of the tangents ratios :${\frac {a}{\sin A}}\,=\,{\frac {b}{\sin B}}\,=\,{\frac {c}{\sin C}}\$ imply the equality of tangents ratios

${\frac {a}{\tan A}}\,=\,{\frac {b}{\tan B}}\,=\,{\frac {c}{\tan C}}\$?--82.137.12.8 (talk) 11:10, 7 May 2018 (UTC)

No, because $\tan \theta ={\frac {\sin \theta }{\cos \theta }}$, but if you really want, then we can rewrite sin as $\sin \theta =\cos \theta \tan \theta$, which you can substitute in to the sine rule equations. Iffy★Chat -- 11:35, 7 May 2018 (UTC)

Points on sides of a triangle being the vertices of an equilateral triangle

How must three points M, N, Q be placed on each of the sides of a triangle ABC having different lengths of sides a, b, c in order that these points be the vertices of an equilateral triangle MNQ? Is an answer to this problem possible by using Heron's formula?(Thanks.)--82.137.12.247 (talk) 13:52, 6 May 2018 (UTC)

There may be many choices possible... In special case a=b=c, you can choose M to divide AB in arbitrary ratio, then define N and Q to divide BC and CA, respectively, in the same ratio. --CiaPan (talk) 16:49, 6 May 2018 (UTC)

This is an image I found at math.stackexchange.com: https://i.stack.imgur.com/Q4qWO.gif

And here is an animated example at GeoGebra: https://www.geogebra.org/m/ey3aYYkK You can drag vertices A, B, C and D to see what happens. --CiaPan (talk) 18:56, 6 May 2018 (UTC)

Here's one solution. Bisect an angle and find whee the bisector cuts the opposite side. Draw lines at 30 degrees back towards towards the angle on either side of that bisecting line. They will hit the other two sides to form the other two corners of an equilateral triangle. Dmcq (talk) 18:43, 6 May 2018 (UTC)

In fact it is even easier. Just mark a point on one side, rotate the whole triangle 60 degrees about that point and where the other two lines intersect forms another point of the equilateral triangle. Dmcq (talk) 20:21, 6 May 2018 (UTC)

These are a bit hard to follow; could you explain in a bit more detail? It makes sense that there will be one degree of freedom in the solution since you have three d.o.f in the selection of the points and essentially only two constraints. --RDBury (talk) 07:21, 7 May 2018 (UTC)

Are there really three d.o.f in selecting points when the initial triangle is scalene thus having lengths of sides a, b, c not equal? Can some relations be found between ratios of segments determined on the three sides? Can the above section problem involving sines ratios be used for determining ratios of segments and angles when the initial scalene triangle is divided in four triangles, one being the equilateral one?--82.137.13.158 (talk) 11:47, 7 May 2018 (UTC)

Suppose the original triangle is ABC going counterclockwise and X is on BC. Rotate the triangle ABC 60 degrees anticlockwise around the point X to A'B'C'. The line C'A' will intersect the line AB at Z say. Then line ZX is one side of the equilateral triangle and Y the point on CA can be got by rotating Z around X to intersect CA. This complex number construction of Loraof below is this geometric construction turned into algebra. Dmcq (talk) 11:44, 10 May 2018 (UTC)

The first construction I gave is quite different, it only gives one equilateral triangle for each of the three angles rather than being a general solution. Dmcq (talk) 11:53, 10 May 2018 (UTC)

Here’s a general algebraic approach, with vertices considered as points a+bi in the complex plane. Scale and position the triangle ABC so it has vertices at A(0,0), B(1,0),and C(p,q). We impose four conditions: M must be on AC, whose equation is b=(q/p)a; N must be on AB, whose equation is b=0; Q must be on BC, whose equation is b=[q/(p–1)](a–1); and MNQ must be equilateral, which by Equilateral triangle#Other properties holds if and only if $M+\left(-1+{\frac {\sqrt {3}}{2}}i\right)N+\left(-1-{\frac {\sqrt {3}}{2}}i\right)Q=0.$ Write $M=M_{a}+M_{b}i$ and likewise for N and Q, and insert these into the four conditions above. This gives five equations since the equilateral condition is really two conditions: the real part of the left side equals 0, and the imaginary part of the left side equals zero. Thus we have five equations in six unknowns (the real and imaginary parts of each of M, N, and Q). With one excess degree of freedom, five of these can be solved for in terms of the sixth one. However, not all solutions will be valid because in some cases one of the found points may not be on the original triangle’s side, but rather just on its extended side. Loraof (talk) 17:32, 7 May 2018 (UTC)

Can the euclidean distance be used instead of complex numbers?--82.137.15.131 (talk) 20:35, 9 May 2018 (UTC)

I think you meant to write $M+\left({\frac {-1+{\sqrt {3}}i}{2}}\right)N+\left({\frac {-1-{\sqrt {3}}i}{2}}\right)Q=0$ but otherwise this looks like the right way to approach the problem; the problem is reduced to a problem in linear algebra. --RDBury (talk) 16:08, 9 May 2018 (UTC)

Right—thanks for catching that. Loraof (talk) 18:12, 9 May 2018 (UTC)

It seems to be a locus problem, by the simulation with GeoGebra. What is the extremum(minimum?) of the length of the side of the equilateral triangle as a function of lengths a, b, c excluding the extended sides cases?--82.137.15.131 (talk) 20:35, 9 May 2018 (UTC)

Since the problem is essentially linear, including the extended sides amounts to adding inequality constraints to the variables and this turns the problem from linear algebra to linear programming. This would imply that the extrema (triangles with max and min side) must occur where one of the vertices of the equilateral triangle are one of the vertices of the original triangle. I'm guessing the details would devolve into a bunch of tedious cases though.

On further reflection it seems that this is malarkey; the length of the sides of the triangle isn't linear so linear programming doesn't apply. There's only one degree of freedom though, so if you can find the side as a function of a single parameter you might be able to apply calculus, though the expressions are probably too hairy to handle without computer algebra. --RDBury (talk) 11:03, 10 May 2018 (UTC)

Based on Loraof's observation, here is a purely geometric construction. First, assume wlog that ABC are in counter-clockwise order. (If not then either relabel or reverse the orientation of the following.) Pick X on AB and Y and AC, then construct Z so that XZY is an equilateral triangle orientated counter-clockwise. If Z is in the interior of ABC then move X and Y further away from A, extending AB and/or AC if necessary. Since Z is outside ABC, AZ must cut BC, at Q say. Construct M on AB so AM:MX = AQ:QZ and N on AC so AN:NY = AQ:QZ. The third point of the equilateral triangle based on MN will be Q, which follows immediately from the linearity of the problem but should be straightforward to prove geometrically using similar triangles. What's going on here is that with X and Y chosen so that XY is parallel to a fixed line l, the locus of the points Z is a line through A and the figures AXZY are all similar. Once you've constructed one such figure it's a matter of finding the appropriate scaling factor so that Z lies on BC. The choice of the direction of l gives you the 1 d.o.f. in the solution. --RDBury (talk) 04:42, 10 May 2018 (UTC)

Well, actually the problem is analytically tractable. It can be solved in the following way. The initial triangle is specified by one side $x_{0}$ and two angles $\alpha$ and $\beta$. The inner unilateral triangle has side $a$.(see figure)

Quite obviously

$\delta +\gamma ={\frac {2\pi }{3}}$

Using sine theorem

${\frac {a}{\sin \alpha }}={\frac {x_{1}}{\sin(\pi -\alpha -\gamma )}}$

${\frac {a}{\sin \beta }}={\frac {x_{0}-x_{1}}{\sin(\pi -\beta -\delta )}}$

Excluding $a$ and $\delta$

$\tan \gamma ={\frac {\sin \alpha \sin(2\pi /3+\beta )-\sin \beta \sin \alpha (x_{0}/x_{1}-1)}{\sin \alpha \cos(2\pi /3+\beta )+\sin \beta \cos \alpha (x_{0}/x_{1}-1)}}$

$a={\frac {\sin \alpha {\sqrt {1+\tan ^{2}\gamma }}}{\sin \alpha +\cos \alpha \tan \gamma }}x_{1}$

They depend on one parameter $x_{1}$. In a special case when the initial triangle is unilateral itself

$\tan \gamma ={\sqrt {3}}{\frac {x_{0}-x_{1}}{3x_{1}-x_{0}}}$

$a={\sqrt {3}}{\frac {\sqrt {1+\tan ^{2}\gamma }}{{\sqrt {3}}+\tan \gamma }}x_{1}$

for example, when $x_{1}=x_{0}/2$ it follows that $\tan \gamma ={\sqrt {3}}$, $a=x_{0}/2$ and $\gamma =\delta =\pi /3$, which is expected. Ruslik_Zero 18:51, 11 May 2018 (UTC)

I like this approach and the result isn't as bad as I thought it would be, but the OP was looking for max and min and I still wouldn't want to try find a critical point of a as a function of x₁. Btw, if angles B and C are both < π/3 then there is a way to inscribe an equilateral triangle in ABC with a vertex at A and the other two on BC. This is more or less excluded by the way the problem was worded, but it seemed worth mentioning. --RDBury (talk) 22:28, 11 May 2018 (UTC)