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# 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.