The Bevan Point

Today we shall look at something different from usual, and turn our attention to a famous triangle center, named after Benjamin Bevan — the circumcenter of the excentral triangle or the Bevan point $Be$ , which is Kimberling triangle center $X_{40}$ . A reasonable amount of work (mainly analytical) has been done regarding this point during the early 20th century, although it has not yet been featured extensively in mathematical olympiads.

Notation will comply with the point definitions in the above diagram, to which a clickable GeoGebra link is attached.

A summary of the key results pertaining to this configuration,

First observe that since and are the orthocenter and nine-point center of the excentral triangle respectively it is well known that the Bevan point is the reflection of the incenter of across its circumcenter.
- Returning to the definition, since in intouch and excentral triangles are homothetic, it follows that their center of homothety lies on the line $\overline{OBe}$ which solves 2003 Vietnam TST Problem 2.
- Also, quadrilateral $L_aIN_aBe$ has diagonals which bisect each other, implying that it is a parallelogram. Since $N_a$ is the midpoint of segment $II_a$ we also have that quadrilateral $L_aN_aI_aBe$ is a parallelogram.
- Due to this reflection, it follows that $S_aA \perp ABe$ where $S_a$ is the $A-$ Sharkydevil point of $\triangle ABC$ , which is the crucial claim for 2013 ELMO Shortlist G2.
- Since it is well known that the circumcenter and orthocenter are isogonal conjugates, $Be$ is the isogonal conjugate of $I$ in the excentral triangle and hence, $I_aBe \perp BC$ and similarly. Shifting our reference triangle from the excentral triangle to $ABC$ we immediately have 2015 Macedonia MO Problem 1.
- Finally, if $N_a,N_b,N_c$ are the minor arc midpoints in $(ABC)$ , by the Incenter-Excenter lemma we know that $N_a$ is the midpoint of segment $II_a$ and similarly. A homothety centered at $I$ with scale factor 2 now implies that the reflection of $I$ across the midpoint $M_a$ of $BC$ lies on $\overline{I_aBe}$ and similarly, which disposes of this problem by betongblander.
Denote by and the , and antipodes in . Since it follows that so by the Mean Geometry theorem the midpoint of segment lies on the perpendicular bisector of segment .
- A homothety centered at $I$ with scale factor 2 sends this perpendicular bisector to the line $\overline{L_aBeA'}$ proving the collinearity of these lines.
- Further, quadrilaterals $AX_bX_cBe$ and similarly are cyclic due to the right angles so their centers $O_a,O_b,O_c$ are the midpoints of segments $ABe$ and similarly. The previous observation then implies that the lines through $O_a$ parallel to the internal $\angle BAC-$ bisector and similarly all pass through the circumcenter $O$ , which solves 2023 Bulgaria MO Problem 2.
- Since $BCB'C'$ is a rectangle and $L_a$ is equidistant to $B$ and $C$ it follows that $L_a$ is the $B'C'$ arc midpoint which contains $A$ . Hence, $\overline{A'L_a}$ is the internal $\angle B'A'C'-$ bisector which in conjunction with the previous observation implies that $Be$ is the incenter of $\triangle A'B'C'$ .
- Moreover, a homothety centered at $H$ with scale factor 2 maps the medial triangle to the antipodal triangle and hence the Spieker center $Sp$ to the Bevan point $Be$ , proving the well known fact that $Sp$ is the midpoint of segment $HBe$ , which has appeared in a contest from Bulgaria in 2004.
- Further, a straightforward angle chase shows that quadrilaterals $ABeA'I_a$ and similarly are all concyclic.
Since the diagonals of quadrilateral bisect each other at where is the de Longchamps point of , it must be a parallelogram. Since we proved previously that is the midpoint of and it is well known that is the midpoint of , it follows that is also a parallelogram. So, points and are collinear.
- It is well known that points $I,G,Sp$ and $Na$ all lie on the Nagel line of $\triangle ABC$ with $IG:GN = 1:2$ and $IG:GSp = 2:1$ (homothety centered at the centroid $G$ mapping the reference triangle to the medial triangle) so $(NaG;SpI)=-1$ . Projecting through $H$ it follows that $Be$ is the midpoint of $NaL$ .
Returning to the reflections of across the midpoints and , observe that passes through . A homothety centered at with scale factor 2 maps this to the circle .
- A homothety at $I$ with scale factor $\frac12$ followed by a homothety at $C$ with scale factor 2 maps $\triangle CE_aE_b$ to $\triangle AIB$ which is well known to be similar to $\triangle CN_aN_b$ . Hence it is clear that $\triangle CE_aE_b \sim \triangle CN_aN_b$ .
- Thus, $C$ is the center of spiral similarity mapping segment $N_aN_b$ to $E_aE_b$ and hence also the center of spiral similarity mapping segment $N_aE_a$ to $N_bE_b$ . Spiral similarity properties imply that $N_aE_a \cap N_bE_b$ lies on $(ABC)$ . Applying this on all three sides it follows that $N_aE_a,N_bE_b$ and $N_cE_c$ concur on $(ABC)$ , which is the statement of 2020 International Festival of Young Mathematicians Grade 10-12 Problem 6.
- With a bit more work, one can also show that this concurrence point is the orthopole of $\overline{IO}$ with respect to the excentral triangle.
Let and denote the intersections of with such that is on the same side of as . Since it is well known that is the reflection of across the midpoint of segment , it follows that is an isosceles trapezoid. In particular, this implies that is the antipode in .
- Hence, a homothety centered at $I$ with scale factor $\frac12$ maps $\overline{TBe}$ to the perpendicular bisector of segment $TI$ , so it follows that quadrilateral $L_aITN_a$ is an isosceles trapezoid.
- It is well known that points $L_a,D$ and $E_o$ are collinear where $E_o$ denotes the $A-$ Evan is Old point. Hence, it is not hard to see that quadrilateral $L_aTE_oBe'$ is concyclic by Power of a point at $D$ .
- Now, an easy angle chase shows that points $A,E_o$ and $Be'$ are collinear, and since it is well known that the homothety center of the intouch and excentral triangles, $X_{57}$ lies on the Evan is Old cevians, $X_{57}$ also lies on this line which settles Problem 95 from i3435’s Muricaaa.
Note that where is the foot of the altitude lies on by the radical center theorem.
- It is not hard to see due to isogonality and rtbc inversion that $AIA' \sim API_a$ . Then an easy angle chase shows that points $Q,I$ and $A'$ are collinear.
- A further angle chase utilizing this similarity shows that $QI_aA'A$ is cyclic, which implies that $Q$ also lies on the circle established in the second bullet point, which is a central claim to both 2018 ELMO Shortlist G5 and 2024 IGO Advanced Problem 3.
Several applications of the Pythagorean theorem show that the Bevan point satisfies the length condition $ABe^2 + X_aBe^2 = BBe^2 + X_bBe^2 = CBe^2 + X_cBe^2$ , which appears in 2003 IMO Shortlist G3 which claims that the Bevan point is the unique point which satisfies this condition with respect to it’s pedal triangle.
The Darboux cubic is known to pass through the points , , the orthocenter , the incenter the circumcenter , the infinity point along the -altitude , the de Longchamps point (which is also the pivot), and the Bevan point and since is the pivot, through the point .
- By Cayley-Bacharach on $AH\infty_A$ , $BeOI$ and $BLX$ , it follows that points $BBeIAX\infty_A$ are coconic.
- By Pascal’s theorem on the coconic hexagon $BBe\infty_aIXA$ it follows that points $ID \cap AB$ , $BBe \cap IX$ and the intersection of the reflection of $ID$ across the midpoint of $BC$ and $AC$ are all collinear, solving this interesting problem from Twitch Solves ISL by DottedCaculator.