This is a paragraph.

Concyclic with the orthocenter I

We illustrate how basic alterations to the classic altitude definition result in concurrent lines, in such a way that the points of concurrency are conclyclic with the orthocenter. Similar constructions applied to the medial triangle yield points that are concyclic with the circumcenter.

Extra points

Our intention at the beginning of the year was to focus most of our attention on four points N,P,W,R associated with any triangle, depicted below:

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By March, much has changed, and we’re grateful to be able to add seven more points, bringing the total, as at this paragraph, to eleven:

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Actually, we have fourteen more points, so that the total, as at the time of this writing, is twenty five. However, these other fourteen points behave totally differently from the above mentioned eleven.

Extended period

Following the addition of twenty one extra points, our initial one-year time frame has been affected. Please help us do the math: “If it takes one year to discuss four points, how long will it take to discuss twenty five points, assuming it takes the same length of time for each point”?

Exchanging pleasantries

Welcome to our theory!

Extensive properties

With reference to the preceding diagram, we have the following individual and collective properties of the points:

  • each point is a point of concurrency of some lines (in this post we show how to obtain W; in the future, we’ll feature how to obtain the rest)
  • R and V are on the circumcircle of the parent \triangle ABC (in fact, RV is a diameter)
  • S and W are on the nine-point circle of the parent \triangle ABC (in fact, SW is a diameter)
  • the line segments KS,OL,UH,VX always make an angle of 45^{\circ} with the positive x-axis
  • the line segments KO,SL,UV,HX always make an angle of 135^{\circ} with the positive x-axis
  • quadrilaterals SKOL and HUVX are always rectangles (follows from the previous two)
  • the two rectangles SKOL and HUVX are “parallel”
  • the area of the bigger rectangle HUVX is four times the area of the smaller rectangle SKOL
  • W is the center of the cyclic quadrilateral HUVX
  • T is the center of the cyclic quadrilateral SKOL
  • rectangles SKOL and HUVX can become squares in special cases (in particular, the extremely pleasant case of a triangle having two sides with reciprocal slopes and one side parallel to the x-axis, and in the expectedly peculiar case of an equilateral triangle)
  • if points R,H,V are connnected to form \triangle RHV, then the Euler line HO is just a median in this triangle
  • if the legs of a right triangle are parallel to the coordinate axes, then rectangle SKOL becomes just the circumcenter and rectangle HUVX becomes the orthocenter (this is a degenerate case, but it does generate additional characterizations of this particular right triangle)
  • if the slopes of the sides of the parent \triangle ABC form a geometric progression, then rectangles SKOL and HUVX “sit” on the same line (the case of geometric progressions is the exquisite platform for the present concept)
  • our favourite point is the wandering point W (this isn’t a property though, but there’s a basis for our bias)
  • And lots more! (this isn’t a property either, but as we explore these points in future posts, more will be shown).

essentially peculiar

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Normal simplifications in the case of equilateral triangles. Squares are formed.

extremely pleasant

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Nice setting: triangles with slopes of the form 0,m,\frac{1}{m}. Squares are also formed as in the case of equilateral triangles.

exquisite platform

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Naturally superior to others. Nicest setting is when the slopes form a geometric progression.

equal points

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No shape is formed, because the points coincide. Still, the triangle more than makes up for it with several characterizations.

elongated path

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Nearly similar to the previous example. The rectangles are stretched out and degenerate to a straight line.

edge partners

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Neighbours sharing edges. A special case of a right triangle with legs’ slopes \pm 1, leading to several points coinciding.

Elementary proofs

Using an idea from our last post, we’ll now define the three lines that have point W as a point of concurrency.
More importantly, you might also observe how we alter the regular slope requirements, and then go ahead of us to define and determine your very own points of concurrency, and then have that “discovery delight”.

Let A(x_1,y_1), B(x_2,y_2), C(x_3,y_3) be the coordinates of the vertices of \triangle ABC, and let m_1, m_2, m_3 be the slopes of sides AB,BC,CA in that order.

Define lines from A,B,C as follows: the line from A has slope \frac{m_1m_3}{-m_2}, the line from B has slope \frac{m_1m_2}{-m_3}, and the line from C has slope \frac{m_2m_3}{-m_1}.

PROVE that the three lines defined above are concurrent.

Consider two such lines through A and B.

(1)   \begin{equation*} \begin{split} y-y_1&=\frac{m_1m_3}{-m_2}(x-x_1)\\ y-y_2&=\frac{m_1m_2}{-m_3}(x-x_2)\\ \end{split} \end{equation*}

The solution to the linear system (1) is:


Let’s verify concurrency. We’ll prove that the solution above satisfies the equation of the line through C:

(2)   \begin{equation*} y-y_3=\frac{m_2m_3}{-m_1}(x-x_3) \end{equation*}

Note that


Consider the left member of equation (2):

    \begin{equation*} \begin{split} y-y_3&=\left(\frac{m_2y_1-m_3y_2}{m_2-m_3}\right)-y_3\\ &=\frac{m_2m_3}{m_2-m_3}(x_1-x_2)\\ \end{split} \end{equation*}

Now consider the right member of equation (2):

    \begin{equation*} \begin{split} \frac{m_2m_3}{-m_1}(x-x_3)&=\frac{m_2m_3}{-m_1}\left(\frac{m_2x_2-m_3x_1}{m_2-m_3}-x_3\right)\\ &=\frac{m_2m_3}{m_2-m_3}\left(\frac{y_2-y_1}{m_2-m_3}\right)\\ &=\frac{m_2m_3}{m_2-m_3}\frac{y_1-y_2}{m_1}\\ &=\frac{m_2m_3}{m_2-m_3}(x_1-x_2) \end{split} \end{equation*}

Admittedly, we skipped a couple of steps, probably because we didn’t want to focus on the details of the algebra, but instead on how the lines were defined. Now go ahead and alter our definition to obtain other points of concurrency.

Given \triangle ABC with vertices at A(1,4), B(0,0), C(3,6), the point W(-1,8) is on its nine-point circle. Find the equations of the three lines through A,B,C that intersect at W.

The slopes of AB, BC,CA are 4,2,1. Set


and define lines through A,B,C with slopes -2,-8,-\frac{1}{2}, respectively:

    \begin{equation*} \begin{split} \textrm{through}~A:~y-4&=-2(x-1)\implies y=-2x+6\\ \textrm{through}~B:~y-0&=-8(x-0)\implies y=-8x\\ \textrm{through}~C:~y-6&=-\frac{1}{2}(x-3)\implies y=-\frac{1}{2}x+\frac{15}{2}\\ \end{split} \end{equation*}

These three equations are all satisfied by x=-1,~y=8.

Given \triangle ABC with vertices at A(1,4), B(0,0), C(3,6), define lines through the midpoints of AB,BC,CA in such a way that the slopes are negatives of the slopes of the sides. PROVE that the three lines meet at the point W(-1,8).

This is actually an interesting property. Basically, we can obtain the point W(-1,8) in two different ways:

  • through each vertex, draw lines with slopes of the form \frac{m_im_j}{-m_k}, where m_k is the slope of the side opposite the reference vertex
  • through each midpoint, draw lines with slopes that are negatives of the slopes of the sides on which the midpoints lie.

In the present case, the slopes of AB, BC,CA are 4,2,1, and the midpoints are \left(\frac{1}{2},2\right), \left(\frac{3}{2},3\right), and (2,5), respectively. Through each of these midpoints we draw lines with slopes -4,-2,-1:

    \begin{equation*} \begin{split} \textrm{through the midpoint of}~AB:~y-2&=-4\left(x-\frac{1}{2}\right)\implies \scriptstyle y=-4x+4\\ \textrm{through the midpoint of}~BC:~y-3&=-2\left(x-\frac{3}{2}\right)\implies \scriptstyle y=-2x+6\\ \textrm{through the midpoint of}~CA:~y-5&=-1(x-2)\implies \scriptstyle y=-x+7\\ \end{split} \end{equation*}

These three equations are all satisfied by x=-1,~y=8.

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The point W on the nine-point circle of \triangle ABC.

Eventful pi

Our final example is a big digression from the preceding discussion. For your pi day celebration.

Let a,b,c be the side-lengths, and let N be the nine-point center, of non-right \triangle ABC. If a=b and AN^2+BN^2+CN^2=\frac{11}{16}c^2, PROVE that \left(\frac{c}{a}\right)^2=\frac{22}{7}.

And so a different version of “\pi” for your pi day celebration.

Let R be the circumradius. Since

    \[AN^2+BN^2+CN^2=\frac{1}{4}(3R^2+a^2+b^2+c^2)\quad\textrm{and}\quad R=\frac{c}{2\sin C},\]

we have \frac{11}{16}c^2=\frac{1}{4}(3R^2+a^2+b^2+c^2). Further simplifications:

    \begin{equation*} \begin{split} 11c^2&=\scriptstyle 12R^2+4a^2+4b^2+4c^2\\ 7c^2-4a^2-4b^2&=12\left(\frac{c}{2\sin C}\right)^2\\ \sin^2C\left(7c^2-4a^2-4b^2\right)&=3c^2\\ (1-\cos^2 C)\Big(7(a^2+b^2-2ab\cos C)-4a^2-4b^2\Big)&=\scriptstyle 3(a^2+b^2-2ab\cos C)\\ \cos C\Big(14ab\cos^2 C-(3a^2+3b^2)\cos C-8ab\Big)&=0\\ 0,\frac{(3a^2+3b^2)\pm\sqrt{(3a^2+3b^2)^2+448a^2b^2}}{28ab}&=\cos C\\ \end{split} \end{equation*}

Since \triangle ABC is not a right triangle (right triangles automatically satisfy the relation AN^2+BN^2+CN^2=\frac{11}{16}c^2 if c is taken as the hypotenuse), we ignore the \cos C=0 value. Then put a=b and consider the negative discriminant (the positive part would give \cos C\geq 1, unacceptable):

    \begin{equation*} \begin{split} \cos C&=\frac{(3a^2+3b^2)-\sqrt{(3a^2+3b^2)^2+448a^2b^2}}{28ab}\\ &=\frac{6a^2-\sqrt{484a^2}}{28a^2}\\ \cos C&=-\frac{4}{7} \end{split} \end{equation*}

By the cosine law again:

    \begin{equation*} \begin{split} c^2&=a^2+b^2-2ab\cos C\\ &=2a^2-2a^2\left(-\frac{4}{7}\right)\\ \therefore c^2&=\frac{22}{7}a^2\\ \left(\frac{c}{a}\right)^2&=\frac{22}{7}. \end{split} \end{equation*}

There \pi goes.

If we didn’t require a=b above, we get some other interesting possibilities. For example, if we want \cos C=-\frac{1}{2}, then \frac{a}{b}=\phi^2 or \left(-\frac{1}{\phi}\right)^2, where \phi is the golden ratio.


The following statements are equivalent for any \triangle ABC:

  1. W=H, where H is the orthocenter
  2. two sides of the triangle are parallel to the coordinate axes
  3. W coincides with one of the triangle’s vertices
  4. the slopes of the three medians form a geometric progression with common ratio r=-2
  5. the slopes of the three medians form an arithmetic progression with common difference equal to -\frac{3}{2} times the first term (d=-\frac{3}{2}a)
  6. one side of the triangle is parallel to the x-axis and another side and the median to it have opposite slopes.

You’re already familiar with the above six equivalent statements. By the time we finish discussing our new points, the number of characterizations will be about twenty, if not more. Be looking forward to that time.


  1. Given \triangle ABC with vertices A(x_1,y_1), B(x_2,y_2), C(x_3,y_3), and m_1,m_2,m_3 as the slopes of sides AB,BC,CA, respectively, PROVE that:
    • \frac{m_2x_2-m_3x_1}{m_2-m_3}=\frac{x_1y_1(x_3-x_2)+x_2y_2(x_1-x_3)+x_3y_3(x_2-x_1)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}
    • \frac{m_2y_1-m_3y_2}{m_2-m_3}=\frac{x_1y_1(y_2-y_3)+x_2y_2(y_3-y_1)+x_3y_3(y_1-y_2)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}
  2. PROVE that the line segments from W to the three vertices and the three midpoints either all make acute angles with the positive x-axis, or all make obtuse angles with the positive x-axis.
  3. Let \cos C^{+}=\frac{(3a^2+3b^2)+\sqrt{(3a^2+3b^2)^2+448a^2b^2}}{28ab}, and let \cos C^{-}=\frac{(3a^2+3b^2)-\sqrt{(3a^2+3b^2)^2+448a^2b^2}}{28ab}. PROVE that:
    • \cos C^{+}\geq 1 (hence there’s no solution except C^{+}=0)
    • -1\leq \cos C^{-}\leq 1.
  4. Let \cos C^{-}=\frac{(3a^2+3b^2)-\sqrt{(3a^2+3b^2)^2+448a^2b^2}}{28ab} as before.
    • If \cos C^{-}=-\frac{1}{2}, PROVE that a^2-3ab+b^2=0
    • Under the above condition, deduce that \frac{a}{b}=\phi^2, or \frac{a}{b}=\left(-\frac{1}{\phi}\right)^2, where \phi is the golden ratio.
  5. Let \cos C^{-}=\frac{(3a^2+3b^2)-\sqrt{(3a^2+3b^2)^2+448a^2b^2}}{28ab} as before.
    • If \cos C^{-}=-\frac{1}{7}, PROVE that a^2-18ab+b^2=0
    • Under the above condition, deduce that \frac{a}{b}=(2\phi+1)^2, or \frac{a}{b}=\left(3-2\phi\right)^2, where \phi is the golden ratio.