## A new point on the nine-point circle II

Take vertex of triangle , together with the orthocenter of the triangle, and our new point with coordinates

(1)

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

## Rigid transversals

If we draw three line segments from to each of the three vertices of any triangle , then the line segments either all make acute angles with the -axis or all make obtuse angles with the -axis. See example 2 below.

In , let be the slopes of sides , respectively. PROVE that the slopes of the line segments are , , , in that order.

Let the coordinates of the vertices be , , and . With as in equation (1), we can calculate the slope of (those of and are similar).

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

In any , PROVE that the line segments either all make acute angles with the positive -axis or all make obtuse angles with the positive -axis.

The slopes from to are given by:

(2)

Re-write the terms in (2) as

(3)

and consider four cases:

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

In , let be the slopes of sides , respectively. PROVE that the line segment is perpendicular to side if and only if .

can also be perpendicular to or , but requiring it to be perpendicular to is more convenient.

First suppose that is perpendicular to . Since the slope of is and the slope of is , we have:

Conversely, suppose that . Let’s simplify the slope of :

This shows that is perpendicular to .

(The case where and is a special case. The case where and is an extremely pleasant case.)

In , let be the slopes of sides , respectively. PROVE that is parallel to if and only if (in other words, the slopes of sides and are negatives of each other).

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

However, since we can’t have in a triangle, we take .

Conversely, suppose that and re-calculate the slopes of and :

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

Let be the orthocenter of , and let be the point with coordinates in equation (1). PROVE that the slope of is the negative reciprocal of the product of the slopes of sides .

If has vertices at , , , then the coordinates of its orthocenter are given by:

(4)

These are very much the same as the coordinates of point , except for the extra term in the numerator of the -coordinate and the extra term in the -coordinate. It follows that the slope of the line segment is:

(5)

where are the slopes of sides .

#### (Main goal)

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

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

Let the slopes of sides be . Assume all non-zero.

The slope from vertex to the orthocenter is (by the definition of the altitude from vertex ). The slope of the line segment is (by example 1), and the slope of the line segment is (by example 5). Since

the slopes of sides form a geometric progression.

## Right triangles

In , let . PROVE that the slope of is the reciprocal of the slope of the hypotenuse .

Let be the slopes of sides . By example 1, the slope of the line segment is . Since the given triangle is right-angled at , sides and are perpendicular, and so , which in turn simplifies the slope of the line segment :

Alternatively, one can also deduce this from example 5.

(The three triangles 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 be a right triangle in which the hypotenuse has slope and the legs are not parallel to the coordinate axes. PROVE that the quadrilateral is a trapezium.

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

Suppose we have and . According to example 7, the slope of will be as well, meaning that is parallel to . This gives the required trapezium . Similarly, if the slope of the hypotenuse is (that is ), then the slope of is again , and this gives a trapezium .

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

Calculate the coordinates of point for with vertices at , , . Deduce that is a trapezium.

Observe that is right-angled at . Using equation (1) with , , and , we get:

Thus, is located at in the Cartesian plane. The slope of is , and the slope of is , so the two line segments are parallel. The slope of is , while the slope of is . Therefore, we obtain a trapezium.

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

Calculate the coordinates of for with vertices at , , and . Deduce that is a trapezium. Verify also that the slopes of the sides of triangles , , and form geometric progressions.

Unlike in the preceding example, the given triangle 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 , , and , we get:

Thus, is the point . With this, the slope of is . Since the slope of is also , it follows that is parallel to . The slope of is and the slope of is . So we obtain a trapezium .

Behind the scences, we calculated the orthocenter of triangle ; it is the point . Together with and the given vertices , , , we have the following slopes:

The slopes of the sides of are then ; the slopes of the sides of are ; the slopes of the sides of are . These form geometric progressions with common ratios , respectively.

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

## Takeaway

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

• the slopes of sides and are reciprocals of each other
• the line segment is perpendicular to side
• is the midpoint of the line segment .

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

1. For with vertices at , , , calculate the slopes of line segments , , and . Verify that is perpendicular to .
2. Let be such that side is parallel to the -axis. PROVE that:
• coincides with the foot of the altitude from vertex
• none of the triangles has slopes in geometric progression.
(This is the only exception to the main point in today’s discussion.)
3. In , let be as given in equation (1). PROVE that the slopes of and are reciprocals of each other if and only if the slope of side is
4. PROVE if the slopes of the sides of a triangle form a geometric progression with common ratio , then the -coordinate of point coincides with the -coordinate of the centroid of the triangle.
5. In , let be the slopes of sides . PROVE that the following statements are equivalent:
• is parallel to

• (Under this condition, the slopes of line segments form a geometric progression.)

## An application of Stewart’s theorem

Let be the nine-point center of triangle . We showed in our previous post that if the parent triangle is right-angled with , then

(1)

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

(2)

being the circumradius and the usual side-lengths.

## Memory recall

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

then we have:

(3)

## Main results

In with the usual notation, PROVE that the orthocenter-circumcenter distance satisfies .

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 (measured from a vertex) and also divides the Euler line in the ratio (measured from the orthocenter).

Consider the diagram below where is the centriod of , is the midpoint of , and (circumradius):

Notice our new point just andering along side . More importantly, observe that is perpendicular to ( is the midpoint of and is the circumcenter). Therefore, the Pythagorean theorem applied to gives:

(4)

Since and , Stewart’s theorem applied to gives:

(5)

The length of median satisfies . Thus, combining equations (4) and (5) gives:

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

Let be the orthocenter of with circum-radius . PROVE that .

Apply Stewart’s theorem to in the diagram below:

Add all the three last equations:

Let and be the nine-point center and orthocenter of . PROVE that .

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

In above, is a median, so:

Similarly, if we connect the segment to vertices and instead, we obtain (with and as medians)

Add the expressions for and simplify:

Since , we have . In turn:

PROVE that , where is the nine-point center of . Hence deduce that in a right triangle with hypotenuse of length , the sum of the squares of the distances from the vertices to the nine-point center is the length of the hypotenuse.

We obtained as an intermediate step in example 3 above.

Now if is such that is the hypotenuse with length , then . The circumradius is half of the length of the hypotenuse, so . Substituting in the expression for , we obtain:

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

Consider in which and . PROVE that .

(If you’re curious as to how we obtained , see example 7 for a sample procedure.)

A key ingredient in the proof is the fact that the circumradius can be expressed as as per the extended law of sines.

Since , the cosine law gives

Also:

Equation (2) becomes:

Show that the circumradius of an equilateral triangle can be given by , where is the length of one of the sides.

The orthocenter and circumcenter coincide for an equilateral triangle, so the distance . Using example 1 and , we have:

In , suppose that . PROVE that either or . If one has in addition, deduce that the triangle is equilateral.

One can easily check that if a right triangle satisfies , then . So the above example is basically saying that the equation is not exclusive to right triangles.

From example 2: . Set the left side equal to :

Thus, either (giving a right triangle), or

as required. Now let , then:

If , then , which is not allowed for a normal non-degenerate triangle. Therefore we must take , which yields . Since we already assumed that , 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 whose coordinates are given by:

(6)

Let be the nine-point center of and let be the point given in equation (6). PROVE that , if has two sides parallel to the coordinate axes.

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

If has two sides parallel to the coordinate axes, PROVE that , where is the point given by equation (6).

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

Let be such that two sides have slopes . PROVE that , where 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 , point coincides with the circumcenter. Therefore we have by example 1 above.

## Takeaway

Let be the side-lengths of , and let be the orthocenter, circumcenter, and nine-point center, respectively.

The following statements are equivalent:

• or

The following statements are also equivalent:

• or

And the following two statements too are equivalent:

• or

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

1. Let be the centroid of , and let be the circumcenter.
• PROVE that , where are the usual side-lengths.
• Deduce that .
2. Let and be the orthocenter and nine-point center of . PROVE that .
3. Let be the orthocenter, circumcenter and nine-point center of . PROVE that:
• .
4. Let be the orthocenter of with circumradius and side-lengths .
• PROVE that
• Deduce that the distance from vertex to the orthocenter is twice the distance from the circumcenter to the midpoint of side .
5. This exercise shows that there are many non-right triangles with the property that the length of the Euler line is half of the length of a side of the triangle. Suppose that in one has . PROVE that:
• or
• if in addition, then . (Thus, any isosceles triangle in which and will satisfy . And it’s even possible to realize this with a non-right, scalene triangle.)
6. Consider with vertices at , , . Verify that:
• the orthocenter is and the circumcenter is
7. Find coordinates for the vertices of a non-right triangle with orthocenter and side-lengths such that .
8. Find coordinates for the vertices of a non-right triangle with nine-point center and side-lengths such that .
9. Consider a quadrilateral with vertices at , , , . PROVE that:
• sides and have opposite slopes, and their lengths are in the ratio
• sides and have opposite slopes, and their lengths are in the ratio
• diagonals and have opposite slopes, and their lengths are in the ratio
• the longer diagonal bisects the shorter diagonal
• the shorter diagonal divides the longer diagonal in the ratio