POLYA022: Triangle centers constructible from O, H, and I (proposed by Paul Yiu, 11/30/01).

This is Problem E2407 of the American Mathematical Monthly, proposed by A.W.Walker:

Given the circumcenter O, the orthocenter H, and the incenter I of an unknown triangle (T),
(a) locate by euclidean construction the Gergonne point and the Lemoine point of (T),
(b) locate the orthocenters of the pedal triangles of H and I.

Monthly editorial Note: This problem is interesting because triangle (T) cannot in general be constructed from the given points, but many points related to (T), including those mentioned in this problem, can be so constructed. The two solutions received are quite involved, so we do not take the space here to print either of them.

Discussion. [JPE, 12/9/01]: We can construct immediately
- the centroid G : Vect(OG) = 1/3 Vect(OH)
- the de Longchamps point L (reflection of H wrt O)
- the Nagel point n (anticomplement of I)
- the NP-center E (midpoint of OH)
Now, as the Feuerbach point F is the center of the rectangular circumhyperbola through I,H,N, we get F as a common point of the NP-circle
of IHN and the line IE ( the second common point is the reflection of I wrt E)
Note that we can construct the incircle (center I going through F) and the circumcircle (center 0 going through X104 = reflection of H wrt F)
Of course, we can construct the Gergonne point g as the second common point of the line IL and the Feuerbach hyperbola.
But we can proceed this way :
The internal center of similitude X_55 of the incircle and the circumcircle is OI inter GF
The external center of similitude X_56 of the incircle and the circumcircle is OI inter HF
The Shiffler point X_21 is nX_55 inter OH
The Gergonne point g is X_56X_21 inter IL
The orthocenter of the pedal triangle of I is ng inter IO.
It remains to find K and the orthocenter of the orthic triangle, but it should not be very difficult.
Note that we get an easy conic construction of ABC because they are the common points - except X_104 _ of the Feuerbach hyperbola and the circumcircle.

[PY, 12/9/01]: This is fantastic! It makes use of Poncelet's theorem that the center of a rectangular hyperbola through 3 given points lies on the nine-point circle of the triangle formed by these 3 points.  So now we know the Feuerbach point. From this we can construct the incircle and the circumcircle. (The circumradius is twice the distance between the Feuerbach point and the nine-point center).
The Gergonne and Nagel points are on the Feuerbach hyperbola since their isogonal conjugates are the centers of similitudes of the circumcircle and the incircle.
Knowing R and r, now we can also determine the point X(57) = (a/(s-a):...:...)  which divides OI in the ratio  OX(57) : OI = 2R+r : 2R-r.

[JPE, 12/9/01]: I read in Clark Kimberling's ETC that the lines L_X74, E_K, H_X52 intersect at X68, which is the isogonal of X24 on the Euler line.
(X74 is the isogonal conjugate of the infinite point of the Euler line and X52 is the orthocenter of the orthic triangle).
As X65 is the isogonal conjugate of X21 (on the Euler line), O,H,K,X65,X68,X74 lie on the Jerabek hyperbola J.
This leads to a construction of K and X52.
Construct the center X125 of J as a common point of the NP-circles of ABC and OHX65 (there are, of course, two common points, but the right one is obvious when we draw the figure) Draw the asymptots of J as the lines going joining X125 and the common points of OH and the NP-circle of ABC.
Now, if a point M lies on a hyperbola W, D is a line going through M, U,U' are the common points of D and the asymptots of W,  the second common point of D and W is the reflection of M wrt the  midpoint of UU'.
Thus we get X74 = reflection of H wrt X125
X68 = second common point of J and the line L_X74
K = second common point of J and the line E_X68
X52 = OK inter H_X68.

[JPE, 12/10/01]: I've just discovered a way to avoid the problem of choice between two points when we try to construct K.
The line through the antipode of the Feuerbach point on the NPC and the complement of the Schiffler point intersects again the NPC at the center X125 of Jerabek hyperbola. (you can get the Schiffler point as the common point of the Euler line and the line joining the Nagel point and the internal center of similitude of the incircle and the circumcircle).
The asymptots of the hyperbola are the lines joining X125 to the common points of the NPC and the Euler line.
Now, draw X74 as the reflection of H wrt X125.
The line Longchamps-X74 intersects again the hyperbola at X68.
The line NPcenter-X68 intersects again the hyperbola at K.
Of course, when we have the asymptotes of an hyperbola, we have immediately the second common point of a given line with the hyperbola if we know the first one.

[JPE, 12/10/01]:  Here is a better construction of the asymptots of the Jerabek hyperbola (we don't need the NP-center of OHX65 and the choice between the two common points; in fact, we don't need X65).
The line through the antipode of F on the NP-circle of ABC and the complement of the Schiffler point intersects again the NP-circle at X125.
The Euler line OH intersects the NP-circle at P, P'; the asymptots are X125_P and X125_P'.

[PY, 12/10/01]: What triangle centers can be constructed from O, H, I?
Based on your use of the Poncelet theorem and the wonderful lemma on second intersection of a line with a rectangular hyperbola, I can do up to X(25) in Kimberling's list, plus a few more. Below I give some details.
Points on the Euler line: O, H, G, L (deLongchamps point), E (nine-point center).
Points on the OI line: centers of similitude of (O) and (I).
Nagel point: superior of I.
The Spieker point X(10) as the inferior of I.
The Feuerbach point is the center of the Feuerbach hyperbola through I, H, N. It lies on the nine-point circle of I, H, N, and can be determined as the intersection of this circle and the line IE.
The incircle and the circumcircle can now be constructed.
The asymptotes of the Feuerbach hyperbola can be constructed.
From these we can constructed the Gergonne point, as the intersection of  IL and the Feuerbach hyperbola.
Now, X(9) = Mittenpunkt = (2,7)/\(4,10).
X(12) = outer Feuerbach point is the harmonic associate of 11 with respect to 1 and 5.
From the diagrams in [TCCT, pp.84,85], we also have the following points on the circumcircle:
X(104) = intersection with (4,11) = reflection of orthocenter in Feuerbach point.
X(100) = antipode of X(104), or on (11,2)
X(105) = intersection with (2,11),
X(106) = intersection with (1,100),
X(101) = intersection with (1,105).
Now, the symmedian point K = X(6) is the intersection  (1,9)/\(101,106).
Up to now we have the first 12 points in ETC.
You have already found X(65) = orthocenter of intouch triangle = OI/\(7,8), and noted that it was on the Jerabek hyperbola. This was one of the points sought in the Monthly problem.
From this, we have the Clawson point: X(19) = (4,9)/\(6,65).
We also know the Jerabek center X(125) from three points on the Jerabek hyperbola H, O and X(65).
X(74) = (4,65)/\circumcircle,
X(107) = (4,74)/\circumcircle,
Now we turn to the Kiepert hyperbola.
The Kiepert center X(115) can be determined from H, G, and the Spieker point.
From the Brocard axis OK, we can also construct the two asymptotes.
Tarry point X(98) = (4,115)/\ circumcircle
Steiner point X(99) = antipode of Tarry point
Kiepert focus X(110) = (centroid, Tarry)/\circumcircle
Parry point X(111) = (centroid, Steiner) /\ circumcircle.
Now, [TCCT] listed a number of central circles through the Parry point.
The isodynamic points X(15), X(16) are the intersection of the Parry circle (2,110,111) and the Brocard axis OK. The two points are inverses with respect to the circumcircle, and X(15) is the interior one.
Now, the Fermat points: X(13) = (2,16)/\(6,115), X(14) =(2,15)/\(6,115),
and the Napoleon points: X(17) = (3,13)/\(4,15) and X(18) = (3,14)/\(4,16).
Now we have everything (in ETC) up to the deLongchamps point. Actually, you already got
the Schiffler point X(21) = OH /\ (8,55).
From [TCCT], we also have
X(22) = Euler line /\ circcle(5,6,111),
X(23) = Euler line /\  circle (3,99,111),
X(25) = Euler line /\  circle (4,6,111),
X(68) = (5,6)/\(20,74) on the Jerabek hyperbola
X(54) = second intersection of (2,68) with Jerabek hyperbola
X(24) = Euler line /\(6,54).
...

[JPE,12/10/01]:  May be, we can try an hazardous conjecture: The triangle centers constructible from O,H, I are exactly those which are constructible from ABC.

[PY, 12/10/01]: Edward Brisse has just sent me some tables of triangle centers. From these we have the simpler constructions:
(i) the Mittenpunkt X(9) can be obtained as the inferior of the Gergonne point;
(ii) X(23) on the Euler line is simply the inversive image of the centroid in the circumcircle.

[Frans Gremen (University of Nijmegen, The Netherlands), 12/4/02]:     Here I'll present some ideas about it. I'm not complete about whether Jean-Pierre' conjecture is a true statement or not.
I start with a possible known fact, that the incenter I cannot be constructed in the OHK-problem. For this problem is given O (the center of the circumcircle), H (the orthocenter) and K (the Lemoine
point). The giving of the sidelengths of triangle OHK is equivalent with  giving R (radius of circumcircle), DELTA (area of triangle ABC) and Q = a^2 + b^2 + c^2 (where a, b and c are the sidelengths of triangle ABC). One has with s = (a + b + c)/2 and an arbitrary x,

(x - s)*(x - a + s)*(x - b + s)*(x - c + s) = x^4 + (-1/2*a^2 - 1/2*b^2 - 1/2*c^2)*x^2 - a*b*c*x + (a^4 + b^4 + c^4 - 2*a^2*b^2 - 2*b^2*c^2 - 2*c^2*a^2)
= x^4 - Q*x^2/2 - 4*R*DELTA x - DELTA^2
s is a root of the the quartic equation in x
x^4 - Q*x^2/2 - 4*R*DELTA x - DELTA^2  = 0            (1)

For Q = 1, DELTA = 2 and R = 3, the equation (1) has a Galois group of S_4. This means s is not square-root expressible in the coefficients of the quartic equation or equivalently in R, Q and DELTA, which are the knowns in the OHK problem. This contradicts the constructibility of I in the OHK  problem, because if I could be constructed then also s could be  constructed.

Before switching to the OHI problem I'll put the last statement in other words by taking sqrt(a) instead of a, also with b and c: The symmetric quantity sqrt(a) + sqrt(b) + sqrt(c) cannot  be equal to an expression involving only the elementary symmetrics
a + b + c, a*b + b*c + c*a and a*b*c using +,-, *, ,/,sqrt.   (2)

Hereafter it is about the OHI problem. From ABC can be constructed a wide class of points, which satisfy the conditions needed in order to be a triangle center, e.g. the point P with barycentric coordinates
(sqrt(a) : sqrt(b) : sqrt(c)). The power of P w.r.t. circumcircle of ABC is equal to

-a^2*sqrt(b*c) - b^2*sqrt(c*a) - c^2*sqrt(a*b)
----------------------------------------------      (3)
(sqrt(a) + sqrt(b) + sqrt(c))^2

On the other hand it is OP^2 - R^2, which would be square-root expressible as in (2) if Jean-Pierre's conjecture is true, which says our P is constructible from O, H and I. Then the more complicated expression in (3) is square-root expressible, but we found in (2) the simpler sqrt(a) + sqrt(b) + sqrt(c) is not. This is still possible, but my doubts are growing. I'am tempting to go on and try to see if P is a counterexample of the conjecture. For this is needed to find the high degree equation with a root the value in (3) and coefficients, which are polynomials in the three quantities
a + b + c, a*b + b*c + c*a and a*b*c.
Substitute in that high degree equation for the latter three quantities some integers. Find finally the order of the Galois group. If it is not a power of 2 then P constitutes a counterexample. Experts in finding the Galois group are the creators of the free downloadable program PARI at the university of Bordeaux (Prof. H. Cohen).

[JPE, 12/5/02]:  Using your method, I think that the conjecture is wrong : your point  P is not constructible from O,H,I.
I choose IH =6; IO = 9; OH = 13; a,b,c are the roots of
x^3-16/65*root(417)*root(65)x^2+29344/65 x-20736/4225*root(417)*root(65) = 0
The power z of P wrt the circumcircle is root of the equation
z^4-51881472/447785 z^3-66788748288/29106025 z^2 +10399537299456/29106025 z + 653965220748460032/122972955625 = 0
and the corresponding Galois group is S4.

Note that we have always
R^2 = IO^2/(2IH^2+2IO^2-OH^2)
a+b+c = root(2IH^2-8IO^2-3OH^2+27R^2)
ab+bc+ca = IH^2-OH^2-4IO^2+9R^2
abc=(a+b+c)IO^2(OH^2-IO^2-2IH^2)/(2IH^2+2IO^2-OH^2)
and that your high degree equation is always of degree 4.

Bibliography. [1] A.W.Walker, Problem E2407, American Mathematical Monthly, 80 (1973) 316; editorial note, 81 (1974) 406.
[TCCT] C. Kimberling, Triangle Centers and Central Triangles, Congressus Numerantium, 129 (1998) 1--295.