Properties of a line

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High School Math Grades 9-12, Ages 13-18

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Properties of a line

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#1 h Mar 5, 2024, 1:54 PM

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If we connect a triangle's incenter and circumcenter, what are the properties of this line?

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#2 h Mar 5, 2024, 3:07 PM

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I think the distance, $d$ between the incentre and circumcentre is
$d = \sqrt{R^2 - 2Rr}$ and that these two centres lie on the Euler line if and only if the triangle is isosceles (including equilateral).

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#3 h Yesterday at 1:51 PM

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Cathong0324 wrote:

If we connect a triangle's incenter and circumcenter, what are the properties of this line?

This is a rather interesting question. I'd like to use this post to talk about some properties of the line $OI$ .
Note: request moving this post to High School Olympiads.
Firstly, for $\triangle{ABC}$ , $ID$ , $IE$ , $IF$ is perpendicular to $BC$ , $CA$ and $AB$ , respectively, at $D$ , $E$ , and $F$ , whereas $G$ , $H$ and $J$ denotes the midpoints of segment $BC$ , $CA$ and $AB$ respectively. $I_A$ , $I_B$ and $I_C$ would be the three escenters. $Cn$ , $Or$ , $Sm$ , $Ge$ and $Ng$ denotes the centroid, orthocenter, symmedian point , gregonne point and nagel point of $\triangle{ABC}$ .

Claim 1 The isogonal conjugate point of $Ge$ is on line $OI$ .
Proof. Since for some of you, this may be well known, as $Ge$ (or $X_7$ )'s isogonal conjugate point is the insimilicenter of $\odot I$ and $\odot O$ , we would do this via barycentrics. The barycentrics of $Ge$ is $\frac1{b+c-a}:\frac1{c+a-b}:\frac1{a+b-c}$ , so the barycentrics of $X_{55}$ would be $a^2(b+c-a):b^2(c+a-b):c^2(a+b-c)$ . The barycentrics of $I$ is $a:b:c$ , and the barycentrics of $O$ is $a^2(b^2+c^2-a^2):b^2(c^2+a^2-b^2):c^2(a^2+b^2-c^2)$ , so the claim is proven.

Claim 2 The circumcenter of $\triangle{I_AI_BI_C}$ is on line $OI$ .
Proof. This is actually very basic. It is obvious that $I_AA\perp{I_BI_C}$ , $I_BB\perp{I_CI_A}$ , $I_CC\perp{I_AI_B}$ . Therefore, $\odot O$ is the nine-point circle of $\triangle I_AI_BI_C$ , $O$ is $\triangle I_AI_BI_C$ 's nine-point center, and $I$ is $\triangle I_AI_BI_C$ 's orthocenter. Therefore, $OI$ is the euler line of $\triangle I_AI_BI_C$ . Prove complete.
Similarly, the orthocenter of $\triangle{DEF}$ would also be on line $OI$ .

(when I was investigating in Novenber last year, I found another interesting fact that I'd like to share.)
Claim 3 The euler line of the arctic triangle of $\triangle{ABC}$ (The triangle formed by the intersections of perpendicular bisectors of two points with the circle on the same side of the circle as the third point, the name of which our classmates gave, and whose official name is 1st circumperp) ( $\triangle A'B'C'$ ) is line $OI$ .
Proof. To be honest, the orthocenter of $\triangle A'B'C'$ is the bevan point of $\triangle ABC$ , and its centroid is $X_{165}$ . Its incenter and nine-point center are also known points, $X_{12518}$ and $X_{3579}$ . We would be using trilinears this time, and trilinears of the vertices of the arctic triangle would be $a:c-b :b-c$ . By basic calculation, we can denote that the centroid of this triangle is the point with trilinears $3a^2-2a(b+c)-(b-c)^2:3b^2-2b(c+a)-(c-a)^2:3c^2-2c(a+b)-(a-b)^2$ . Clearly, it is $-4O+I$ . And since $\triangle A'B'C'$ and $\triangle ABC$ share a circumcenter, $OI$ is the euler line of $\triangle A'B'C'$ .
(By the way, it is easy to prove that the same goes for the antarctica triangle, or 2nd circumperp.)

Now a rather interesting fact about points on line $OI$ :

Claim 4 Point $P$ is on line $OI$ . $PH_A$ , $PH_B$ , $PH_C$ are perpendicular to $BC$ , $CA$ and $AB$ at $H_A$ , $H_B$ , $H_C$ , then $BH_A+CH_B+AH_C=CH_A+AH_B+BH_C$ (note that for example, if $H_A$ is on ray $CB$ , but not segment $CB$ , then in this expression, $BH_A$ is negative).
Proof. This is not hard. We need only say that the when $P$ is $O$ or $I$ , the proof is obvious, and then prove that $\frac{DH_A}{GH_A}=\frac{EH_B}{HH_A}=\frac{FH_C}{JH_C}$ .

There are many lines that are parallel or perpendicularto $OI$ and many points that lie on it. Here are a few examples:

Claim 5 $X$ and $Y$ lie on rays $BA$ and $CA$ , such that $BX=CY=BC$ . Prove that $XY\perp OI$ .
Proof. There are quite a few ways of proving this. However, here is my quickest way.
Line $CI$ intersects $\odot O$ at point $Q\neq C$ . As known to all, $QB=QA=QI$ , $BX\perp OQ$ , $BY\perp CQ$ . Therefore, $\angle XBY=\angle OQI$ , $\frac{OQ}{IQ}=\frac{OQ}{BQ}=\frac{CY}{BY}=\frac{BX}{BY}$ . Therefore, $\triangle{XBY}\sim\triangle{OQI}$ . Since $BX\perp OQ$ , we have $XY\perp OI$ .

More would be filled later.

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