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April 13

Euler line forms a Fixed Angle with a Given Side

We are given a fixed triangle side BC, and asked to find the locus described by a third point A, so that the Euler line of the resulting triangle ABC forms a given angle with side BC. I have solved the problem for 0° and 90° using GeoGebra, but am unable to solve for the general case. (For 90°, the locus of A is -of course- the perpendicular bisector of segment BC, and for 0° we have an ellipse whose minor axis is the segment BC, and whose major axis is BC√3 ). — 79.113.236.232 (talk) 19:50, 13 April 2017 (UTC)[reply]

Here's an idea that will certainly work: coordinatize in your favorite coordinate system, so that B and C are fixed and A has variable coordinates. Choose two points on the Euler line, say the centroid and cicumcenter; give their coordinates. This determines the equation of the Euler line, hence its slope. Solving for a fixed value of the slope gives an implicit equation for the coordinates of A. The solution set is your locus. It will always be some conic section. --JBL (talk) 20:41, 13 April 2017 (UTC)[reply]

Euler line#Slope gives the slope of the Euler line in terms of the slopes of the sides as

m_{E}=-{\frac {m_{1}m_{2}+m_{1}m_{3}+m_{2}m_{3}+3}{m_{1}+m_{2}+m_{3}+3m_{1}m_{2}m_{3}}}.

Set the slope of side BC equal to 0, and set the slope of the Euler line at a given value (the tangent of the required angle). This gives one equation in the two unknown remaining slopes, so one of the remaining slopes can be solved for parametric on the other one. Loraof (talk) 21:53, 13 April 2017 (UTC)[reply]

The questioner says that for an angle of 0°, we have an ellipse whose minor axis is the segment BC, and whose major axis is BC√3 . I would note that this ellipse must be punctured at the two endpoints of the major axis. If B=(–1, 0) and C=(1, 0), so BC=2, the upper end of the (vertical) major axis is at A= (0, ${\sqrt {3}}$ ) (since from the origin to A is half the major axis, the latter being given by the questioner as $2{\sqrt {3}}$ ). Then AB=AC=2, and we have an equilateral triangle, which does not have an Euler line since all the points that would be on the Euler line coincide with each other. Loraof (talk) 17:51, 14 April 2017 (UTC)[reply]

Quite right. I saw no point in adding extra clutter. — 79.113.215.18 (talk) 05:43, 15 April 2017 (UTC)[reply]

For an angle α below 60° I get an ellipse whose center coincides with that of the segment BC, and whose minor axis forms an angle of α⁄2 with the same segment. However, I am unable to obtain a general formula for the lengths of its two axes. For α above 60°, the locus becomes a hyperbola, and for α = 60°, we have two parallel lines. — 79.113.215.18 (talk) 05:43, 15 April 2017 (UTC)[reply]

@79.113.215.18:, @79.113.236.232: Maybe this will work. The equilateral triangle with horizontal base BC of length 1 has m_2=0, m_1 =sqrt(3), and m_3= –sqrt(3). By the slope formula I gave above, the Euler slope is given as 0/0. I'm guessing that this means the equilateral triangle is a limiting case of any of your ellipse points as the x coordinate goes to the midpoint of the side BC. If that speculation is correct, that gives us two points on the ellipse regardless of the required Euler slope (the points above and below the BC axis giving an equilateral triangle). Next, consider a triangle with point A very near the BC axis. In the slope formula, m_2=0 and m_3 approx.=0, so by the formula m_1 = –3/m_E which is strictly positive. So as A goes to the BC axis it cannot be approaching any point other than B or C (lest the slope m_1=0). So by symmetry, the ellipse (regardless of the Euler slope) must cross the BC axis at B and at C.

Now we have four points on any ellipse: B, C, and the two points giving an equilateral triangle. Is that what you found with GeoGebra? If so, these four pieces of information along with the line of the minor axis (slope alpha/2 and going through the midpoint of BC) should be enough to let you specify the ellipse. Loraof (talk) 17:17, 19 April 2017 (UTC)[reply]

Thanks to your observations, I've finally managed to find a fifth point, namely the ellipse's intersection with the vertical axis at (0, ξ m_E). — 86.121.233.7 (talk) 00:26, 20 April 2017 (UTC)[reply]

Resolved