A new point on the nine-point circle II

Take vertex A of triangle ABC, together with the orthocenter H of the triangle, and our new point W with coordinates

(1)   \begin{equation*} \begin{split} x&=\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)}\\ y&=\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)} \end{split} \end{equation*}

then the resulting triangle AHW has the following property: the slopes of its sides form a geometric progression. Same with vertices B and C as well as the corresponding triangles BHW and CHW (examples 6 and 10). This construction works for all triangles, except those triangles with one side parallel to the x or y axis.

Rigid transversals

If we draw three line segments from W to each of the three vertices A,B,C of any triangle ABC, then the line segments AW,BW,CW either all make acute angles with the x-axis or all make obtuse angles with the x-axis. See example 2 below.

In \triangle ABC, let m_1,m_2,m_3 be the slopes of sides AB,BC,CA, respectively. PROVE that the slopes of the line segments AW, BW, CW are \frac{m_1m_3}{-m_2}, \frac{m_1m_2}{-m_3}, \frac{m_2m_3}{-m_1}, in that order.

Let the coordinates of the vertices be A(x_1,y_1), B(x_2,y_2), and C(x_3,y_3). With W as in equation (1), we can calculate the slope of AW (those of BW and CW are similar).

    \begin{equation*} \begin{split} \Delta x&= \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)}-x_1\\ &=\frac{(x_1-x_2)(y_2-y_3)(x_3-x_1)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}\\ \Delta y&=\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)}-y_1\\ &=\frac{(y_1-y_2)(x_2-x_3)(y_1-y_3)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}\\ \therefore \textrm{slope of AW}&=\frac{\Delta y}{\Delta x}\\ &=\frac{(y_1-y_2)(x_2-x_3)(y_1-y_3)}{(x_1-x_2)(y_2-y_3)(x_3-x_1)}\\ &=\frac{y_1-y_2}{x_1-x_2}\times\frac{y_1-y_3}{x_3-x_1}\times\frac{x_2-x_3}{y_2-y_3}\\ &=m_1\times (-m_3)\times\frac{1}{m_2}\\ &=\frac{m_1m_3}{-m_2} \end{split} \end{equation*}

The placement of the negative sign beside m_2 was deliberate: to make it easy to remember the formula for the slope of AW as well as those of BW and CW. Here’s how it works: the slope of AW is the product of the slopes of the two sides (AB and AC) that originate from vertex A divided by the negative of the slope of the side opposite vertex A (side BC).

In any \triangle ABC, PROVE that the line segments AW,BW,CW either all make acute angles with the positive x-axis or all make obtuse angles with the positive x-axis.

The slopes from W to A,B,C are given by:

(2)   \begin{equation*} \frac{m_1m_3}{-m_2},~\frac{m_1m_2}{-m_3},~\frac{m_2m_3}{-m_1} \end{equation*}

Re-write the terms in (2) as

(3)   \begin{equation*} \frac{m_1m_2m_3}{-m_2^2},~\frac{m_1m_2m_3}{-m_3^2},~\frac{m_1m_2m_3}{-m_1^2} \end{equation*}

and consider four cases:

  • Case I: m_1,m_2,m_3 are all positive. In this case, the product m_1m_2m_3 is also positive and so each term in equation (3) is negative.
  • Case II: m_1,m_2,m_3 are all negative. In this case, the product m_1m_2m_3 is also negative and so each term in equation (3) is positive.
  • Case III: two of m_1,m_2,m_3 are positive while the third is negative; specifically, let m_1,m_2 be positive while m_3 is negative. In this case, the product m_1m_2m_3 is negative, and so the terms in equation (3) are all positive.
  • Case IV: one of m_1,m_2,m_3 is positive while the other two are negative; specifically, let m_1 be positive while m_2 and m_3 are negative. In this case, the product m_1m_2m_3 will be positive and so the terms in equation (3) will be negative.

In \triangle ABC, let m_1,m_2,m_3 be the slopes of sides AB,BC,CA, respectively. PROVE that the line segment AW is perpendicular to side BC if and only if m_1m_3=1.

AW can also be perpendicular to AB or AC, but requiring it to be perpendicular to BC is more convenient.

First suppose that AW is perpendicular to BC. Since the slope of AW is \frac{m_1m_3}{-m_2} and the slope of BC is m_2, we have:

    \[\frac{m_1m_3}{-m_2}\times m_2=-1\implies m_1m_3=1.\]

Conversely, suppose that m_1m_3=1. Let’s simplify the slope of AW:

    \[m_{AW}=\frac{m_1m_3}{-m_2}=\frac{1}{-m_2}.\]

This shows that AW is perpendicular to BC.

(The case where m_1m_3=1 and m_2=\pm 1 is a special case. The case where m_1m_3=1 and m_2=0 is an extremely pleasant case.)

In \triangle ABC, let m_1,m_2,m_3 be the slopes of sides AB,BC,CA, respectively. PROVE that AW is parallel to BW if and only if m_2=-m_3 (in other words, the slopes of sides BC and CA are negatives of each other).

First suppose that AW is parallel to BW. Set their slopes equal:

    \[\frac{m_1m_3}{-m_2}=\frac{m_1m_2}{-m_3}\implies m_2^2=m_3^2\implies m_2=\pm m_3.\]

However, since we can’t have m_2=m_3 in a triangle, we take m_2=-m_3.

Conversely, suppose that m_2=-m_3 and re-calculate the slopes of AW and BW:

    \[m_{AW}=\frac{m_1m_3}{-m_2}=\frac{m_1m_3}{m_3}=m_1,~~m_{BW}=\frac{m_1m_2}{-m_3}=\frac{m_1m_2}{m_2}=m_1.\]

What this means is that AW and BW are parallel to side AB (in fact, W is the midpoint of AB under this condition).

Let H be the orthocenter of \triangle ABC, and let W be the point with coordinates in equation (1). PROVE that the slope of HW is the negative reciprocal of the product of the slopes of sides AB,BC,CA.

If \triangle ABC has vertices at A(x_1,y_1), B(x_2,y_2), C(x_3,y_3), then the coordinates of its orthocenter are given by:

(4)   \begin{equation*} \begin{split} x&=\scriptstyle\frac{x_1y_1(x_3-x_2)+x_2y_2(x_1-x_3)+x_3y_3(x_2-x_1)+(y_1-y_2)(y_2-y_3)(y_3-y_1)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}\\ y&=\scriptstyle\frac{x_1y_1(y_2-y_3)+x_2y_2(y_3-y_1)+x_3y_3(y_1-y_2)-(x_1-x_2)(x_2-x_3)(x_3-x_1)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)} \end{split} \end{equation*}

These are very much the same as the coordinates of point W, except for the extra term (y_1-y_2)(y_2-y_3)(y_3-y_1) in the numerator of the x-coordinate and the extra term -(x_1-x_2)(x_2-x_3)(x_3-x_1) in the y-coordinate. It follows that the slope of the line segment HW is:

(5)   \begin{equation*} m_{HW}=\frac{-(x_1-x_2)(x_2-x_3)(x_3-x_1)}{(y_1-y_2)(y_2-y_3)(y_3-y_1)}=\frac{-1}{m_1m_2m_3} \end{equation*}

where m_1,m_2,m_3 are the slopes of sides AB,BC,CA.

(Main goal)

For any triangle ABC with non-zero side-slopes, PROVE that the three triangles AHW, BHW, and CHW have their side-slopes in geometric progressions.

We do this for triangle AHW. The same argument can be adapted for the other two triangles.

Let the slopes of sides AB,BC,CA be m_1,m_2,m_3. Assume all non-zero.

The slope from vertex A to the orthocenter H is \frac{-1}{m_2} (by the definition of the altitude from vertex A). The slope of the line segment WA is \frac{m_1m_3}{-m_2} (by example 1), and the slope of the line segment HW is \frac{-1}{m_1m_2m_3} (by example 5). Since

    \[\frac{m_1m_3}{-m_2}\times\frac{-1}{m_1m_2m_3}=\frac{1}{m_2^2}=\left(\frac{-1}{m_2}\right)^2,\]

the slopes of sides WA,AH,HW form a geometric progression.

Right triangles

In \triangle ABC, let \angle C=90^{\circ}. PROVE that the slope of CW is the reciprocal of the slope of the hypotenuse AB.

Let m_1,m_2,m_3 be the slopes of sides AB,BC,CA. By example 1, the slope of the line segment CW is \frac{m_2m_3}{-m_1}. Since the given triangle is right-angled at C, sides BC and CA are perpendicular, and so m_2m_3=-1, which in turn simplifies the slope of the line segment CW:

    \[\frac{m_2m_3}{-m_1}=\frac{1}{m_1},~m_1\neq 0.\]

Alternatively, one can also deduce this from example 5.

(The three triangles AHW,BHW,CHW reduces to only two for a right triangle, since its orthocenter coincides with a vertex. But we do get a special quadrilateral.)

Related trapezium

Let \triangle ABC be a right triangle in which the hypotenuse has slope \pm 1 and the legs are not parallel to the coordinate axes. PROVE that the quadrilateral ABCW is a trapezium.

Excluding the case where the legs are parallel to the coordinate axes is essential; if not, W coincides with the orthocenter, which is one of the vertices of a right triangle.

Suppose we have \angle C=90^{\circ} and m_{AB}=1. According to example 7, the slope of CW will be 1 as well, meaning that CW is parallel to AB. This gives the required trapezium ABCW. Similarly, if the slope of the hypotenuse AB is -1 (that is m_{AB}=-1), then the slope of CW is again -1, and this gives a trapezium ABCW.

(Note that the slopes of AW and BW are reciprocals of each other under the above condition. Together with the slope of CW as \pm 1, we get that the slopes of the line segments AW,CW,BW form a geometric progression, even though the slopes of the sides of the parent triangle ABC do not form a geometric progression.)

Calculate the coordinates of point W for \triangle ABC with vertices at A(0,0), B(10,10), C(4,12). Deduce that ABCW is a trapezium.

Observe that \triangle ABC is right-angled at C. Using equation (1) with x_1=0,y_1=0, x_2=10,y_2=10, and x_3=4,y_3=12, we get:

    \begin{equation*} \begin{split} x&=\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{0+100(-4)+48(10)}{0+10(12)+4(-10)}\\ &=1\\ y&=\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)}\\ &=\frac{0+100(12)+48(-10)}{0+10(12)+4(-10)}\\ &=9 \end{split} \end{equation*}

Thus, W is located at (1,9) in the Cartesian plane. The slope of CW is \frac{12-9}{4-1}=1, and the slope of AB is 1, so the two line segments are parallel. The slope of AW is 9, while the slope of BC is -\frac{1}{3}. Therefore, we obtain a trapezium.

(Notice that the slopes of BW, CW, AW form a geometric progression, despite the fact the the slopes of the parent triangle ABC do not form a geometric progression.)

Calculate the coordinates of W for \triangle ABC with vertices at A(0,0), B(6,12), and C(5,20). Deduce that AWBC is a trapezium. Verify also that the slopes of the sides of triangles AHW, BHW, and CHW form geometric progressions.

Unlike in the preceding example, the given triangle ABC is not right-angled this time. So this example is saying that we can still obtain this special trapezium even when the starting triangle is not right-angled. Using equation (1) with x_1=0,y_1=0, x_2=6,y_2=12, and x_3=5,y_3=20, we get:

    \begin{equation*} \begin{split} x&=\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{0+72(-5)+100(6)}{0+6(20)+5(-12)}\\ &=4\\ y&=\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)}\\ &=\frac{0+72(20)+100(-12)}{0+6(20)+5(-12)}\\ &=4 \end{split} \end{equation*}

Thus, W is the point (4,4). With this, the slope of BW is \frac{12-4}{6-4}=4. Since the slope of AC is also 4, it follows that AC is parallel to BW. The slope of AW is 1 and the slope of BC is -8. So we obtain a trapezium AWBC.

Behind the scences, we calculated the orthocenter of triangle ABC; it is the point H\left(36,\frac{9}{2}\right). Together with W(4,4) and the given vertices A(0,0), B(6,12), C(5,20), we have the following slopes:

    \[\scriptstyle m_{HW}=\frac{1}{64},~m_{AW}=1,~m_{BW}=4,~m_{CW}=16,~m_{HA}=\frac{1}{8},~m_{HB}=-\frac{1}{4},~m_{HC}=-\frac{1}{2}.\]

The slopes of the sides of \triangle AHW are then \frac{1}{64},\frac{1}{8},1; the slopes of the sides of \triangle BHW are \frac{1}{64},-\frac{1}{4},4; the slopes of the sides of \triangle CHW are \frac{1}{64},-\frac{1}{2},16. These form geometric progressions with common ratios 8,-16,-32, respectively.

(Notice that the slopes of AW, BW, CW form a geometric progression, despite the fact the the slopes of the parent triangle ABC do not form a geometric progression.)

Takeaway

Let W be the point on the nine-point circle of \triangle ABC whose coordinates were given in equation (1). Let H be the orthocenter of \triangle ABC. Then the following statements are equivalent:

  • the slopes of sides AB and AC are reciprocals of each other
  • the line segment WA is perpendicular to side BC
  • W is the midpoint of the line segment AH.

You’ll be getting used to equivalent statements by now.

Tasks

  1. For \triangle ABC with vertices at A(-3,6), B(0,0), C(2,16), calculate the slopes of line segments AW, BW, and CW. Verify that AW is perpendicular to AB.
  2. Let \triangle ABC be such that side BC is parallel to the x-axis. PROVE that:
    • W coincides with the foot of the altitude from vertex A
    • none of the triangles AWH,BWH,CWH has slopes in geometric progression.
      (This is the only exception to the main point in today’s discussion.)
  3. In \triangle ABC, let W be as given in equation (1). PROVE that the slopes of AW and BW are reciprocals of each other if and only if the slope of side AB is \pm 1.
  4. PROVE if the slopes of the sides of a triangle form a geometric progression with common ratio r=-2, then the x-coordinate of point W coincides with the y-coordinate of the centroid of the triangle.
  5. In \triangle ABC, let m_1,m_2,m_3 be the slopes of sides AB,BC,CA. PROVE that the following statements are equivalent:
    • WA is parallel to BC
    • m_2^2+m_1m_3=0
      (Under this condition, the slopes of line segments BW,AW,CW form a geometric progression.)

An application of Stewart’s theorem

Let N be the nine-point center of triangle ABC. We showed in our previous post that if the parent triangle is right-angled with \angle B=90^{\circ}, then

(1)   \begin{equation*}AN^2+BN^2+CN^2=\frac{11}{16}AC^2. \end{equation*}

We’ll now prove what holds more generally for any triangle:

(2)   \begin{equation*}AN^2+BN^2+CN^2=\frac{1}{4}\left(3R^2+a^2+b^2+c^2\right),\end{equation*}

R being the circumradius and a,b,c the usual side-lengths.

Memory recall

Let’s first be reminded of the main ingredient: Stewart’s theorem. It states that given a triangle ABC with side-lengths a,b,c and a cevian AW like the one depicted below:

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then we have:

(3)   \begin{equation*} b^2m+c^2n=a(d^2+mn) \end{equation*}

Main results

In \triangle ABC with the usual notation, PROVE that the orthocenter-circumcenter distance HO satisfies HO^2=9R^2-(a^2+b^2+c^2).

It was from this amazing site that we first learnt of the above identity. We’ll add our own proof of it to an already existing pool of proofs, using Stewart’s theorem as a tool. We’ll also use the formula for median lengths, and the fact that the centroid divides a median in the ratio 2:1 (measured from a vertex) and also divides the Euler line HO in the ratio 2:1 (measured from the orthocenter).

Consider the diagram below where G is the centriod of \triangle ABC, M is the midpoint of BC, and AO=BO=CO=R (circumradius):

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Notice our new point W just Wandering along side BC. More importantly, observe that MO is perpendicular to BC (M is the midpoint of BC and O is the circumcenter). Therefore, the Pythagorean theorem applied to \triangle COM gives:

(4)   \begin{equation*} \begin{split} MO^2&=R^2-\left(\frac{a}{2}\right)^2\\ MO^2&=\frac{1}{4}(4R^2-a^2) \end{split} \end{equation*}

Since AG:GM=2:1 and HG:GO=2:1, Stewart’s theorem applied to \triangle AMO gives:

(5)   \begin{equation*} \begin{split} MO^2(AG)+R^2(GM)&=AM(AG\times GM+GO^2)\\ MO^2\left(\frac{2}{3}AM\right)+R^2\left(\frac{1}{3}AM\right)&=AM\left(\frac{2}{9}AM^2+\frac{1}{9}HO^2\right)\\ 2MO^2+R^2&=\frac{2}{3}AM^2+\frac{1}{3}HO^2 \end{split} \end{equation*}

The length of median AM satisfies AM^2=\frac{2b^2+2c^2-a^2}{4}. Thus, combining equations (4) and (5) gives:

    \begin{equation*} \begin{split} \frac{1}{2}\left(4R^2-a^2\right)+R^2&=\frac{2}{3}\left(\frac{2b^2+2c^2-a^2}{4}\right)+\frac{1}{3}HO^2\\ 3R^2-\frac{1}{2}a^2&=\frac{1}{6}(2b^2+2c^2-a^2)+\frac{1}{3}HO^2\\ 3R^2-\frac{1}{6}(2b^2+2c^2-a^2+3a^2)&=\frac{1}{3}HO^2\\ \therefore 9R^2-(a^2+b^2+c^2)&=HO^2 \end{split} \end{equation*}

The above procedure works regardless of the vertex we started with. And it also works even in the peculiar case where A,H,O are co-linear. It also doesn’t matter if both H and O are outside the parent triangle.

Let H be the orthocenter of \triangle ABC with circum-radius R. PROVE that AH^2+BH^2+CH^2=12R^2-(a^2+b^2+c^2).

Apply Stewart’s theorem to \triangle AHO in the diagram below:

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    \begin{equation*} \begin{split} AH^2\Big(\frac{1}{3}HO\Big)+R^2\Big(\frac{2}{3}HO\Big)&=HO\Big(\frac{4}{9}AM^2+\frac{2}{9}HO^2\Big)\\ AH^2+2R^2&=\frac{4}{3}\Big(\frac{2b^2+2c^2-a^2}{4}\Big)+\frac{2}{3}HO^2\\ \therefore AH^2+2R^2&=\frac{2b^2+2c^2-a^2}{3}+\frac{2}{3}HO^2\\ \therefore BH^2+2R^2&=\frac{2a^2+2c^2-b^2}{3}+\frac{2}{3}HO^2\\ \therefore CH^2+2R^2&=\frac{2a^2+2b^2-c^2}{3}+\frac{2}{3}HO^2 \end{split} \end{equation*}

Add all the three last equations:

    \begin{equation*} \begin{split} AH^2+BH^2+CH^2+6R^2&=a^2+b^2+c^2+2HO^2\\ \therefore AH^2+BH^2+CH^2&=a^2+b^2+c^2+2OH^2-6R^2\\ &=\scriptstyle a^2+b^2+c^2+2\Big(9R^2-(a^2+b^2+c^2)\Big)-6R^2\\ &=12R^2-(a^2+b^2+c^2) \end{split} \end{equation*}

Let N and H be the nine-point center and orthocenter of \triangle ABC. PROVE that AN^2+BN^2+CN^2+HN^2=3R^2.

We only slightly modify the proof given in example 2, replacing the centroid with the nine-point center N (note that N is the midpoint of HO).

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In \triangle AHO above, AN is a median, so:

    \[AN^2=\frac{2AH^2+2R^2-HO^2}{4}.\]

Similarly, if we connect the segment HO to vertices B and C instead, we obtain (with BN and CN as medians)

    \[BN^2=\frac{2BH^2+2R^2-HO^2}{4},~CN^2=\frac{2CH^2+2R^2-HO^2}{4}.\]

Add the expressions for AN^2, BN^2, CN^2 and simplify:

    \begin{equation*} \begin{split} AN^2+BN^2+CN^2&=\frac{1}{4}\left(6R^2+2(AH^2+BH^2+CH^2)-3HO^2\right)\\ &=\scriptstyle\frac{1}{4}\Big(6R^2+2(12R^2-(a^2+b^2+c^2))-3(9R^2-(a^2+b^2+c^2))\Big)\\ &=\frac{1}{4}(3R^2+a^2+b^2+c^2)\\ \end{split} \end{equation*}

Since HN=\frac{1}{2}HO, we have HN^2=\frac{1}{4}HO^2=\frac{1}{4}(9R^2-(a^2+b^2+c^2)). In turn:

    \begin{equation*} \begin{split} AN^2+BN^2+CN^2+HN^2&=\scriptstyle\frac{1}{4}\Big(3R^2+a^2+b^2+c^2+9R^2-(a^2+b^2+c^2)\Big)\\ &=3R^2 \end{split} \end{equation*}

PROVE that AN^2+BN^2+CN^2=\frac{1}{4}(3R^2+a^2+b^2+c^2), where N is the nine-point center of \triangle ABC. Hence deduce that in a right triangle ABC with hypotenuse of length c, the sum of the squares of the distances from the vertices to the nine-point center is 11/16 the length of the hypotenuse.

We obtained AN^2+BN^2+CN^2=\frac{1}{4}(3R^2+a^2+b^2+c^2) as an intermediate step in example 3 above.

Now if \triangle ABC is such that AB is the hypotenuse with length c, then a^2+b^2=c^2. The circumradius R is half of the length of the hypotenuse, so R=\frac{c}{2}. Substituting in the expression for AN^2+BN^2+CN^2, we obtain:

    \begin{equation*} \begin{split} AN^2+BN^2+CN^2&=\frac{1}{4}(3R^2+a^2+b^2+c^2)\\ &=\frac{1}{4}\Big(3\left(\frac{c}{2}\right)^2+c^2+c^2\Big)\\ &=\frac{1}{4}\Big(\frac{11}{4}c^2\Big)\\ &=\frac{11}{16}c^2. \end{split} \end{equation*}

We show in example 5 below that the above relation also holds in a non-right triangle setting.

Consider \triangle ABC in which a=b and \cos C=-\frac{4}{7}. PROVE that AN^2+BN^2+CN^2=\frac{11}{16}c^2.

(If you’re curious as to how we obtained \cos C=-\frac{4}{7}, see example 7 for a sample procedure.)

A key ingredient in the proof is the fact that the circumradius R can be expressed as R=\frac{c}{2\sin C}=\frac{b}{2\sin B}=\frac{a}{2\sin A} as per the extended law of sines.

Since \cos C=-\frac{4}{7}, the cosine law gives

    \[c^2=a^2+b^2-2ab\cos C=a^2+a^2-2a^2\left(-\frac{4}{7}\right)=\frac{22}{7}a^2.\]

Also:

    \[\sin^2 C=1-\cos^2 C=1-\left(-\frac{4}{7}\right)^2=\frac{33}{49}.\]

Equation (2) becomes:

    \begin{equation*} \begin{split} AN^2+BN^2+CN^2&=\frac{1}{4}\left(3R^2+a^2+b^2+c^2\right)\\ &=\frac{1}{4}\left(\frac{3c^2}{4\left(33/49\right)}+\frac{7}{22}c^2+\frac{7}{22}c^2+c^2\right)\\ &=\frac{c^2}{4}\left(\frac{49}{44}+\frac{14}{44}+\frac{14}{44}+\frac{44}{44}\right)\\ &=\frac{c^2}{4}\left(\frac{121}{44}\right)\\ &=\frac{11}{16}c^2. \end{split} \end{equation*}

Show that the circumradius R of an equilateral triangle ABC can be given by R=\frac{a}{\sqrt{3}}, where a is the length of one of the sides.

The orthocenter and circumcenter coincide for an equilateral triangle, so the distance HO=0. Using example 1 and a=b=c, we have:

    \[0=9R^2-(a^2+a^2+a^2)\implies R=\sqrt{\frac{3a^2}{9}}=\frac{a}{\sqrt{3}}.\]

In \triangle ABC, suppose that AH^2+BH^2+CH^2=c^2. PROVE that either \cos C=0 or \cos C=\frac{3a^2+3b^2\pm\sqrt{9(a^2+b^2)^2-32a^2b^2}}{8ab}. If one has a=b in addition, deduce that the triangle is equilateral.

One can easily check that if a right triangle satisfies a^2+b^2=c^2, then AH^2+BH^2+CH^2=c^2. So the above example is basically saying that the equation AH^2+BH^2+CH^2=c^2 is not exclusive to right triangles.

From example 2: AH^2+BH^2+CH^2=12R^2-(a^2+b^2+c^2). Set the left side equal to c^2:

    \begin{equation*} \begin{split} c^2&=12R^2-a^2-b^2-c^2\\ 2c^2+a^2+b^2&=12R^2\\ 2c^2+a^2+b^2&=12\left(\frac{c}{2\sin C}\right)^2\\ 2c^2+a^2+b^2&=\left(\frac{3c^2}{\sin^2 C}\right)\\ \sin^2 C\Big(2c^2+a^2+b^2\Big)&=3c^2\\ \sin^2 C\Big(2(a^2+b^2-2ab\cos C)+a^2+b^2\Big)&=3(a^2+b^2-2ab\cos C)\\ (1-\cos^2 C)\Big(3a^2+3b^2-4ab\cos C\Big)&=3a^2+3b^2-6ab\cos C\\ 4ab\cos^3 C-(3a^2+3b^2)\cos^2 C+2ab\cos C&=0\\ \cos C\Big(4ab\cos^2C-(3a^2+3b^2)\cos C+2ab\Big)&=0 \end{split} \end{equation*}

Thus, either \cos C=0 (giving a right triangle), or

    \[\cos C=\frac{3a^2+3b^2\pm\sqrt{9(a^2+b^2)^2-32a^2b^2}}{8ab},\]

as required. Now let a=b, then:

    \begin{equation*} \begin{split} \cos C&=\frac{6a^2\pm\sqrt{9(2a^2)^2-32a^4}}{8a^2}\\ &=\frac{6a^2\pm\sqrt{4a^4}}{8a^2}\\ &=\frac{6a^2\pm 2a^2}{8a^2}\\ &=1,\frac{1}{2} \end{split} \end{equation*}

If \cos C=1, then \angle C=0^{\circ}, which is not allowed for a normal non-degenerate triangle. Therefore we must take \cos C=\frac{1}{2}, which yields \angle C=60^{\circ}. Since we already assumed that a=b, it follows that we obtain an equilateral triangle.

Mere repetitions

What follows are particular cases of some of the preceding examples, applied to our new point W whose coordinates are given by:

(6)   \begin{equation*} \begin{split} x&=\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)}\\ y&=\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)} \end{split} \end{equation*}

Let N be the nine-point center of \triangle ABC and let W be the point given in equation (6). PROVE that AN^2+BN^2+CN^2+WN^2=3R^2, if \triangle ABC has two sides parallel to the coordinate axes.

Trivial stuff. If \triangle ABC has legs parallel to the coordinate axes, then W=H, based on one of the equivalent conditions in our previous post. The conclusion now follows from example 3.

If \triangle ABC has two sides parallel to the coordinate axes, PROVE that AW^2+BW^2+CW^2=12R^2-(a^2+b^2+c^2), where W is the point given by equation (6).

Trivial stuff, twice. Under the given condition, we have W=H. Thus, the conclusion follows from example 2 above.

Let \triangle ABC be such that two sides have slopes \pm 1. PROVE that HW^2=9R^2-(a^2+b^2+c^2), where W is the point with coordinates given by equation (6).

Trivial stuff, thrice. The given triangle is obviously a right triangle. Since the legs have slopes \pm 1, point W coincides with the circumcenter. Therefore we have HW^2=9R^2-(a^2+b^2+c^2) by example 1 above.

Takeaway

Let a,b,c be the side-lengths of \triangle ABC, and let H,O,N be the orthocenter, circumcenter, and nine-point center, respectively.

The following statements are equivalent:

  • HO^2=\frac{c^2}{4}
  • \cos C=0 or \cos C=\frac{9a^2+9b^2\pm\sqrt{(9a^2+9b^2)^2-320a^2b^2}}{20ab}.

The following statements are also equivalent:

  • AH^2+BH^2+CH^2=c^2
  • \cos C=0 or \cos C=\frac{3a^2+3b^2\pm\sqrt{(3a^2+3b^2)^2-32a^2b^2}}{8ab}.

And the following two statements too are equivalent:

  • AN^2+BN^2+CN^2=\frac{11}{16}c^2
  • \cos C=0 or \cos C=\frac{3a^2+3b^2\pm\sqrt{(3a^2+3b^2)^2+448a^2b^2}}{28ab}.

Bottom line: Many non-right triangles share properties (relating to H,O,N and stuff) that one would have thought to be exclusive to the right triangle.

Tasks

  1. Let G be the centroid of \triangle ABC, and let O be the circumcenter.
    • PROVE that AG^2+BG^2+CG^2=\frac{1}{3}\left(a^2+b^2+c^2\right), where a,b,c are the usual side-lengths.
    • Deduce that HO^2=3(3R^2-AG^2-BG^2-CG^2).
  2. Let H and N be the orthocenter and nine-point center of \triangle ABC. PROVE that AH^2+BH^2+CH^2-HO^2=AN^2+BN^2+CN^2+HN^2.
  3. Let H,O,N be the orthocenter, circumcenter and nine-point center of \triangle ABC. PROVE that:
    • AH^2+BH^2+CH^2=AN^2+BN^2+CN^2+\frac{5}{4}HO^2
    • AH^2+BH^2+CH^2=AN^2+BN^2+CN^2+5HN^2.
  4. Let H be the orthocenter of \triangle ABC with circumradius R and side-lengths a,b,c.
    • PROVE that AH^2=4R^2-a^2
    • Deduce that the distance from vertex A to the orthocenter is twice the distance from the circumcenter to the midpoint of side BC.
  5. This exercise shows that there are many non-right triangles with the property that the length of the Euler line HO is half of the length of a side of the triangle. Suppose that in \triangle ABC one has HO=\frac{c}{2}. PROVE that:
    • \cos C=0 or \cos C=\frac{9a^2+9b^2\pm\sqrt{(9a^2+9b^2)^2-320a^2b^2}}{20ab}
    • if a=b in addition, then \cos C=0,1,\frac{4}{5}. (Thus, any isosceles triangle in which a=b and \cos C=4/5 will satisfy HO=c/2. And it’s even possible to realize this with a non-right, scalene triangle.)
  6. Consider \triangle ABC with vertices at A(0,0), B(4,0), C(2,6). Verify that:
    • the orthocenter is H\left(2,\frac{2}{3}\right) and the circumcenter is O\left(2,\frac{8}{3})
    • HO=\frac{1}{2}AB.
  7. Find coordinates for the vertices of a non-right triangle ABC with orthocenter H and side-lengths a,b,c such that AH^2+BH^2+CH^2=c^2.
  8. Find coordinates for the vertices of a non-right triangle ABC with nine-point center N and side-lengths a,b,c such that AN^2+BN^2+CN^2=\frac{11}{16}c^2.
  9. Consider a quadrilateral ABCD with vertices at A(0,0), B(1,4), C(3,6), D(6,-6). PROVE that:
    • sides AB and CD have opposite slopes, and their lengths are in the ratio 1:3
    • sides AD and BC have opposite slopes, and their lengths are in the ratio 1:3
    • diagonals AC and BD have opposite slopes, and their lengths are in the ratio 3:5
    • the longer diagonal BD bisects the shorter diagonal AC
    • the shorter diagonal AC divides the longer diagonal BD in the ratio 1:9
    • ABCD is a cyclic quadrilateral.
      (The point D that partly facilitated these features is among the quartet of points we’ll introduce later this year, if time/space permits.)
  10. Consider \triangle ABC with vertices at A\left(\frac{p^2-q^2}{p}\right), B(0,0), C(p,q), p\neq q. PROVE that AH^2+BH^2+CH^2=4m_{c}^2, where m_c is the length of the median from vertex C.