# Difference between revisions of "Nine point circle"

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* The three [[midpoint]]s of the [[edge]]s of the triangle. | * The three [[midpoint]]s of the [[edge]]s of the triangle. | ||

* The three midpoints of the segments joining the [[vertex | vertices]] of the triangle to its [[orthocenter]]. (These points are sometimes known as the [[Euler point]]s of the triangle.) | * The three midpoints of the segments joining the [[vertex | vertices]] of the triangle to its [[orthocenter]]. (These points are sometimes known as the [[Euler point]]s of the triangle.) | ||

+ | |||

+ | "The nine-point circle is tangent to the incircle, has a radius equal to half the circumradius, and its center is the midpoint of the segment connecting the orthocenter and the circumcenter." | ||

+ | -hankinjg | ||

That such a circle exists is a non-trivial theorem of Euclidean [[geometry]]. | That such a circle exists is a non-trivial theorem of Euclidean [[geometry]]. | ||

Line 10: | Line 13: | ||

The center of the nine point circle is the [[nine-point center]] and is usually denoted <math>N</math>. | The center of the nine point circle is the [[nine-point center]] and is usually denoted <math>N</math>. | ||

+ | The nine point circle is tangent to the [[incircle]], has a radius equal to half the [[circumradius]], and its center is the midpoint of the segment connecting the orthocenter and the [[circumcenter]], upon which the [[centroid]] also falls. | ||

It's also denoted Kimberling center <math>X_5</math>. | It's also denoted Kimberling center <math>X_5</math>. | ||

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==First Proof of Existence== | ==First Proof of Existence== | ||

− | Since <math>O_c</math> is the midpoint of <math>AB</math> and <math>E_b</math> is the midpoint of <math>BH</math>, <math>O_cE_b</math> is parallel to <math>AH</math>. Using similar logic, we see that <math>O_bE_c</math> is also parallel to <math>AH</math>. Since <math>E_b</math> is the midpoint of <math>HB</math> and <math>E_c</math> is the midpoint of <math> | + | Since <math>O_c</math> is the midpoint of <math>AB</math> and <math>E_b</math> is the midpoint of <math>BH</math>, <math>O_cE_b</math> is parallel to <math>AH</math>. Using similar logic, we see that <math>O_bE_c</math> is also parallel to <math>AH</math>. Since <math>E_b</math> is the midpoint of <math>HB</math> and <math>E_c</math> is the midpoint of <math>HC</math>, <math>E_bE_c</math> is parallel to <math>BC</math>, which is perpendicular to <math>AH</math>. Similar logic gives us that <math>O_bO_c</math> is perpendicular to <math>AH</math> as well. Therefore <math>O_bO_cE_bE_c</math> is a rectangle, which is a cyclic figure. The diagonals <math>O_bE_b</math> and <math>O_cE_c</math> are diagonals of the circumcircle. Similar logic to the above gives us that <math>O_aO_cE_aE_c</math> is a rectangle with a common diagonal to <math>O_bO_cE_bE_c</math>. Therefore the circumcircles of the two rectangles are identical. We can also gain that rectangle <math>O_aO_bE_aE_b</math> is also on the circle. |

We now have a circle with the points <math>O_a</math>, <math>O_b</math>, <math>O_c</math>, <math>E_a</math>, <math>E_b</math>, and <math>E_c</math> on it, with diameters <math>O_aE_A</math>, <math>O_bE_b</math>, and <math>O_cE_c</math>. We now note that <math>\angle E_aH_aO_a=\angle E_bH_bO_b=\angle E_cH_cO_c=90^{\circ}</math>. Therefore <math>H_a</math>, <math>H_b</math>, and <math>H_c</math> are also on the circle. We now have a circle with the midpoints of the sides on it, the three midpoints of the segments joining the vertices of the triangle to its orthocenter on it, and the three feet of the altitudes of the triangle on it. Therefore the nine points are on the circle, and the nine-point circle exists. | We now have a circle with the points <math>O_a</math>, <math>O_b</math>, <math>O_c</math>, <math>E_a</math>, <math>E_b</math>, and <math>E_c</math> on it, with diameters <math>O_aE_A</math>, <math>O_bE_b</math>, and <math>O_cE_c</math>. We now note that <math>\angle E_aH_aO_a=\angle E_bH_bO_b=\angle E_cH_cO_c=90^{\circ}</math>. Therefore <math>H_a</math>, <math>H_b</math>, and <math>H_c</math> are also on the circle. We now have a circle with the midpoints of the sides on it, the three midpoints of the segments joining the vertices of the triangle to its orthocenter on it, and the three feet of the altitudes of the triangle on it. Therefore the nine points are on the circle, and the nine-point circle exists. | ||

==Second Proof of Existence== | ==Second Proof of Existence== | ||

− | We know that the reflection of the orthocenter about the sides and about the midpoints of the triangle's sides lie on the circumcircle. Thus, consider the homothety centered at <math>H</math> with ratio <math> | + | We know that the reflection of the orthocenter about the sides and about the midpoints of the triangle's sides lie on the circumcircle. Thus, consider the homothety centered at <math>H</math> with ratio <math>{1}/{2}</math>. It maps the circumcircle of <math>\triangle ABC</math> to the nine point circle, and the vertices of the triangle to its [[Euler point]]s. |

Hence proved. | Hence proved. | ||

− | |||

− | |||

==See also== | ==See also== | ||

*[[Kimberling center]] | *[[Kimberling center]] |

## Latest revision as of 09:59, 31 May 2020

The **nine point circle** (also known as *Euler's circle* or *Feuerbach's circle*) of a given triangle is a circle which passes through 9 "significant" points:

- The three feet of the altitudes of the triangle.
- The three midpoints of the edges of the triangle.
- The three midpoints of the segments joining the vertices of the triangle to its orthocenter. (These points are sometimes known as the Euler points of the triangle.)

"The nine-point circle is tangent to the incircle, has a radius equal to half the circumradius, and its center is the midpoint of the segment connecting the orthocenter and the circumcenter." -hankinjg

That such a circle exists is a non-trivial theorem of Euclidean geometry.

The center of the nine point circle is the nine-point center and is usually denoted .

The nine point circle is tangent to the incircle, has a radius equal to half the circumradius, and its center is the midpoint of the segment connecting the orthocenter and the circumcenter, upon which the centroid also falls.

It's also denoted Kimberling center .

## First Proof of Existence

Since is the midpoint of and is the midpoint of , is parallel to . Using similar logic, we see that is also parallel to . Since is the midpoint of and is the midpoint of , is parallel to , which is perpendicular to . Similar logic gives us that is perpendicular to as well. Therefore is a rectangle, which is a cyclic figure. The diagonals and are diagonals of the circumcircle. Similar logic to the above gives us that is a rectangle with a common diagonal to . Therefore the circumcircles of the two rectangles are identical. We can also gain that rectangle is also on the circle.

We now have a circle with the points , , , , , and on it, with diameters , , and . We now note that . Therefore , , and are also on the circle. We now have a circle with the midpoints of the sides on it, the three midpoints of the segments joining the vertices of the triangle to its orthocenter on it, and the three feet of the altitudes of the triangle on it. Therefore the nine points are on the circle, and the nine-point circle exists.

## Second Proof of Existence

We know that the reflection of the orthocenter about the sides and about the midpoints of the triangle's sides lie on the circumcircle. Thus, consider the homothety centered at with ratio . It maps the circumcircle of to the nine point circle, and the vertices of the triangle to its Euler points. Hence proved.

## See also

*This article is a stub. Help us out by expanding it.*