Let $ABC$ be a triangle with incenter $I$, and suppose that $AI$, $BI$, and $CI$ intersect $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. Let the circumcircles of $BDF$ and $CDE$ intersect at $D$ and $P$, and let $H$ be the orthocenter of $DEF$. Prove that $HI=HP$
I have been thinking about this geometry test question for a long time, but I haven't made it yet. This question was given to me by a friend of mine. According to him, it was the last question of a contest test
On
(1) The quick computer solution.
This post could only find a simple analytic (but straightforward) solution, using the barycentric coordinates for the three points involved in the problem, which are in terms of the ETC (Encyclopedia of Triangle Centers) $$ \begin{aligned} I &= X(\color{red}1) \\ &= (\ a\ :\ :\ )\ , \\ H &= X(\color{red}{500})\\ &= (\ a^2(b^2 +c^2 - a^2 + bc)[2abc + (b + c)(a^2 - (b - c)^2)] \ :\ :\ )\ , \\ P &= X(\color{red}{501})\\ &= (\ a^2[a^3 - b^3 - c^3 - bc(a + b + c) + ab(a - b) + ac(a - c)]/(b + c) \ :\ :\ )\ , \end{aligned} $$ and the "consecutive points" $500$, $501$ are involved in our problem maybe only by pure coincidence.
These two points are "rather complicated in their nature", and the list of the properties in the ETC show that there is no quick, no simple "common spiritual neighbour" of them, so that a synthetic argument can be applied. The sage code checking the same distance is simple:
R.<a,b,c> = QQ[]
X1 (a, b, c) = a
X500(a, b, c) = a^2 * (b^2 + c^2 - a^2 + b*c) * (2*a*b*c + (b + c)*(a^2 - (b - c)^2))
X501(a, b, c) = a^2 * (a^3 - b^3 - c^3 - b*c*(a + b + c) + a*b*(a - b) + a*c*(a - c)) / (b + c)
I = vector( [ X1(a, b, c), X1(b, c, a), X1(c, a, b)] ) # unnormalized barycentrics
H = vector( [X500(a, b, c), X500(b, c, a), X500(c, a, b)] ) # unnormalized barycentrics
P = vector( [X501(a, b, c), X501(b, c, a), X501(c, a, b)] ) # unnormalized barycentrics
I = 1/(I[0] + I[1] + I[2]) * I # normalized barycentrics
H = 1/(H[0] + H[1] + H[2]) * H # normalized barycentrics
P = 1/(P[0] + P[1] + P[2]) * P # normalized barycentrics
def dist2(P, Q):
x, y, z = list(P-Q)
return -a^2*y*z -b^2*x*z -c^2*x*y
dIH2 = dist2(I,H).factor()
dPH2 = dist2(P,H).factor()
print bool( dIH2 == dPH2 )
resulting in a more or less illuminating
True
and one could claim this is it in this century!
(2.1) Computer solution also computing all involved points. The barycentric coordinates of the involved points.
However, let us insert some more human details and some steps that may make the task digestible for a pure geometer. We will work with barycentric coordinates as it is the case in most problems involving such complicated points. (We will work in the sequel with barycentric rather than trilinear coordinates.)
As a reference for the terminology i will use:
Max Schindler, Evan Chen, Barycentric Coordinates for the Impatient
For a point $P$ (in a general setting, not the one in the problem) we use tacitly the notation $P=(x_P,y_P,z_P)$ for the normalized barycentric coordinates of $P$ w.r.t. an once for all times fixed triangle $\Delta ABC$ with sides $a,b,c$, so that we have: $$ \begin{aligned} P&=x_PA+y_PB+z_PC\ ,\\ 1&=x_P+y_P+z_P\ . \end{aligned} $$ If trilinears are needed, we will tacitly use capital letters, $X_P,Y_P,Z_P$, and the translations are: $$ \begin{aligned} X_P&=x_P/a\ , & Y_P&=y_P/b\, & Z_p &=z_P/c\ ;\\ x_P&=aX_P\ , & y_P&=bY_P\, & z_p &= cZ_P\ ; \\ \end{aligned} $$ $$ \begin{aligned} 1 &=x_P+y_P+z_P\\ &=aX_P+bY_P+cZ_P\ . \end{aligned} $$ In our case, all coordinates of $X(k)$, $k\in 1, 500, 501,\dots $ (where the dots stay for some other relatives of these centers, that may enter the stage,) are rational functions of $a,b,c$. Computations will usually involve homogeneous expressions, so sometimes we use coordinates of the shape $kx_P, ky_P,kz_P$, and we will write $P=(kx_P: ky_P:kz_P)$, $k\ne 0$ suitable scalar.
The (not normalized, i.e. sum is $\ne 0$ but may be $\ne 1$,) coordinates of the points $I;D,E,F$ are rather simple: $$ \begin{aligned} I &=(a:b:c)\ ,\\ D &=(\color{red}0:b:c)\ ,\\ E &=(a:\color{red}0:c)\ ,\\ F &=(a:b:\color{red}0)\ ,\\ \end{aligned} $$ Let now $H$ be the orthocenter of $\Delta DEF$. Then the conditions $DH\perp EF$ and the similar ones obtained cyclically are written algebraically: $$ \begin{aligned} D&= (0:b:c) = \left(0, \frac b{b+c}, \frac c{b+c} \right)\ ,\\ E&= (a:0:c) = \left(\frac a{a+c}, 0, \frac c{a+c} \right)\ ,\\ F&= (a:b:0) = \left(\frac a{a+b}, \frac b{a+b}, 0 \right)\ ,\\[2mm] \overrightarrow{EF} &= \left(\frac a{a+b}, \frac b{a+b}, 0 \right) - \left(\frac a{a+c}, 0, \frac c{a+c} \right) \\ &= \left( \frac {a(c-b)}{(a+b)(a+c)}, \frac b{a+b}, -\frac c{a+c} \right) \\ &=\frac 1{(a+b)(a+c)}(\ a(c-b),\ b(a+c),\ -c(a+b)\ )\ . \\ &\qquad\text{ so the equation of the line $DH$ (through $D$ and $\perp EF$) is} \\ DH:\ 0= &+ a^2\Big((\color{blue}{y_H}-y_D)(z_F-z_E)+(\color{blue}{z_H}-z_D)(y_F-y_E)\Big) \\ & + b^2\Big((\color{blue}{x_H}-x_D)(z_F-z_E)+(\color{blue}{z_H}-z_D)(x_F-x_E)\Big) \\ & + c^2\Big((\color{blue}{x_H}-x_D)(y_F-y_E)+(\color{blue}{y_H}-y_D)(x_F-x_E)\Big) \ . \end{aligned} $$ This is explicitly $$ \boxed{\ \begin{aligned} (\operatorname{eq}_H(a,b,c)) \qquad 0= &-bc(b-c)(b+c)(a+b+c)\color{blue}{x_H} \\ & -ac(b+c)(a+c)(a+b-c)\color{blue}{y_H} \\ &+ab(b+c)(a+b)(a-b+c)\color{blue}{z_H} \\ & +abc(b-c)(a+b+c) \ . \end{aligned} \ } $$ Here, we have used instead of $(x_F-x_E,y_F-y_E,z_F-z_E)$ the version without denominators $(a(c-b)\ :\ b(a+c)\ :\ -c(a+b))$. The coefficient of $\color{blue}{x_H}$ is then instead of $b^2(z_F-z_E)+c^2(y_F-y_E)$ the expression $-b^2\cdot c(a+b)+c^2\cdot b(a+c)=bc(-ab-b^2+ac+c^2)=bc(c-b)(a+(b+c))$. This is still conceivable for a human.
Using trilinears, we have arguably a simpler situation: $$ \begin{aligned} 1= &+(b+c)\color{blue}{X_H} \\ & +\frac{b+c}{b-c}(a+c)\frac{p-c}p\color{blue}{Y_H} \\ &-\frac{b+c}{b-c}(a+b)\frac{p-b}p\color{blue}{Z_H} \ . \end{aligned} $$ The barycentrics of $H$ are thus determined by $\operatorname{eq}_H(a,b,c)$, $\operatorname{eq}_H(b,c,a)$, $\operatorname{eq}_H(c,a,b)$, and the norming $x_H+y_H+z_H=1$.
Let us also get the equations for $P=(x:y:z)$. Consider the circle $(AEF)$, it has a homogenous equation of the shape $$ 0=-a^2yz -b^2xz -c^2xy + (ux+vy+wz)(x+y+z) \ . $$ The point $A(1:0:0)$ satisfies, so $u=0$. Inserting also $F(a:b:0)$ we obtain $-0-0-c^2\; ab+(0+vb+0)(a+b+0)=0$, which determines $v$. Analogously for $E(a:0:c)$ (via $b\leftrightarrow c$), and thus we get the equation $$ \boxed{\ \begin{aligned} \operatorname{eq}_P(a,b,c) \qquad 0 &=-a^2yz -b^2xz -c^2xy \\ &\qquad+ \left(\frac{ac^2}{a+b}y+\frac{ab^2}{a+c}z\right)(x+y+z) \ . \end{aligned} \ } $$ The barycentrics $(x_P:y_P:z_P)$ of $P$ (now with the index $P$ for a short time interval) are thus determined by $\operatorname{eq}_P(a,b,c)$, $\operatorname{eq}_P(b,c,a)$, $\operatorname{eq}_P(c,a,b)$, and the norming $x_P+y_P+z_P=1$. Expanded (and without index): $$ \begin{aligned} a^2yz +b^2xz +c^2xy &=\frac{ac^2}{a+b}y+\frac{ab^2}{a+c}z \\ &=\frac{bc^2}{a+b}x+\frac{a^2b}{b+c}z \\ &=\frac{b^2c}{a+c}x+\frac{a^2c}{b+c}y\ , \\ x+y+z &= 1\ . \end{aligned} $$
Using these formulas, we can check the result in the OP with sage, postpone in (2.4) further human steps.
(2.2) Computer solution also computing all involved points, the code.
Using the knowledge of the equations for $H=X(500)$, $P=X(501)$ from (2.1), we can use sage to obtain the points, and also check the formulas from ETC. For instance:
S.<a,b,c> = PolynomialRing(QQ)
K = S.fraction_field()
R.<xH,yH,zH, xP,yP,zP> = PolynomialRing(K)
xI, yI, zI = a/(a+b+c), b/(a+b+c), c/(a+b+c)
def eqH(a,b,c, xH, yH, zH):
return ( -(b-c)*(b+c)*(a+b+c) * xH/a
-(b+c)*(a+c)*(a+b-c) * yH/b
+(b+c)*(a+b)*(a-b+c) * zH/c
+(b-c)*(a+b+c) )
def eqP(a,b,c, xP,yP,zP):
return ( (-a^2*yP*zP -b^2*xP*zP -c^2*xP*yP) * (a+b)*(a+c)
+ (a*c^2*(a + c)*yP + a*b^2*(a + b)*zP) * (xP+yP+zP) )
eqs = [ xH+yH+zH -1,
xP+yP+zP -1,
eqH(a,b,c, xH,yH,zH),
eqH(b,c,a, yH,zH,xH),
eqH(c,a,b, zH,xH,yH),
eqP(a,b,c, xP,yP,zP),
eqP(b,c,a, yP,zP,xP),
eqP(c,a,b, zP,xP,yP),
]
J = R * eqs # ideal generated by above eqs
dIH2 = ( -a^2*(yI-yH)*(zI-zH)
-b^2*(xI-xH)*(zI-zH)
-c^2*(xI-xH)*(yI-yH) )
dPH2 = ( -a^2*(yP-yH)*(zP-zH)
-b^2*(xP-xH)*(zP-zH)
-c^2*(xP-xH)*(yP-yH) )
dIH2 - dPH2 in J
Which delivers in the sage interpreter after a paste of the above code:
....: dIH2 - dPH2 in J
True
sage:
This is an other computer proof, using the equation of the involved points, rather than directly their coordinates provided by the ETC.
Observation: The code:
sage: for bas in J.groebner_basis():
....: print bas
....: print
shows in fact a basis with elements bas of the shape xH +(rational expression in $a,b,c$), and similar for the other components. This means that algebraically we have a unique solution.
(2.3) Computer check of the formulas from ETC. (This alternative way to program in sage uses solve, getting explicit formulas, in contrast to the above ideal (algebraic geometry) point of view.)
var('a,b,c');
var('x,y,z');
D = vector([0, b/(b+c), c/(b+c)])
E = vector([a/(a+c), 0, c/(a+c)])
F = vector([a/(a+b), b/(a+b), 0])
EF, FD, DE = F-E, D-F, E-D
H = vector([x, y, z])
def eq(P, Q, U, V):
"""Given P, Q and U, V in baricentric coordonated,
deliver as equation the condition for PQ _|_ UV
We use tacitly the globals a, b, c
"""
pq1, pq2, pq3 = Q - P
uv1, uv2, uv3 = V - U
eq = ( 0 == a^2 * ( pq2*uv3 + pq3*uv2 ) + b^2 * ( pq1*uv3 + pq3*uv1 ) + c^2 * ( pq1*uv2 + pq2*uv1 ) )
return eq
eqDH = eq(D, H, E, F)
eqEH = eq(E, H, F, D)
eqFH = eq(F, H, D, E)
sol = solve( [eqDH, eqEH, eqFH, x+y+z == 1], [x, y, z], solution_dict=True )[0]
X = sol[x].factor()
Y = sol[y].factor()
Z = sol[z].factor()
def X_formula(a, b, c):
"""Formula up to factor from
http://faculty.evansville.edu/ck6/encyclopedia/ETC.html
"""
return a^2 * (b^2 +c^2 - a^2 + b*c) * (2*a*b*c + (b + c)*(a^2 - (b - c)^2))
print "Check formula:"
print "Check X / Y ::", bool( X/Y == X_formula(a,b,c) / X_formula(b,c,a) )
print "Check Y / Z ::", bool( Y/Z == X_formula(b,c,a) / X_formula(c,a,b) )
print "Check Z / X ::", bool( Z/X == X_formula(c,a,b) / X_formula(a,b,c) )
This checks the formula for $h$:
Check formula:
Check X / Y :: True
Check Y / Z :: True
Check Z / X :: True
Similarly for $P$, the code:
var('a,b,c');
var('x,y,z');
eq(a, b, c, x, y, z) = \
0 == -a^2*y*z -b^2*x*z -c^2*x*y + (a*c^2/(a + b)*y + a*b^2/(a + c)*z)*(x+y+z)
eq_AEF = eq(a, b, c, x, y, z)
eq_BFD = eq(b, c, a, y, z, x)
eq_CDE = eq(c, a, b, z, x, y)
eq_ONE = x + y + z == 1
sol = solve( [eq_AEF, eq_BFD, eq_CDE, eq_ONE], [x, y, z], solution_dict=True )[0]
P = vector( [sol[x], sol[y], sol[z]] )
def g(a, b, c):
return ( a^2 *
(a^3 - b^3 - c^3 - b*c*(a + b + c)
+ a*b*(a - b) + a*c*(a - c))
/ (b + c) )
print "Check P[0]/P[1] = g(a, b, c) / g(b, c, a) ::", bool( P[0]/P[1] == g(a, b, c) / g(b, c, a) )
print "Check P[1]/P[2] = g(b, c, a) / g(c, a, b) ::", bool( P[1]/P[2] == g(b, c, a) / g(c, a, b) )
print "Check P[2]/P[0] = g(c, a, b) / g(a, b, c) ::", bool( P[2]/P[0] == g(c, a, b) / g(a, b, c) )
This gives:
Check P[0]/P[1] = g(a, b, c) / g(b, c, a) :: True
Check P[1]/P[2] = g(b, c, a) / g(c, a, b) :: True
Check P[2]/P[0] = g(c, a, b) / g(a, b, c) :: True
(2.4) Human manipulations of equations, just an unsucessful attempt. (It shows also why a synthetic proof is hard to finde.)
We want to show the equality of distances $d(H,I)^2=d(H,P)^2$. Algebraically, using cyclic sums $a\to b\to c\to a$, $x\to y\to z\to x$: $$ -\sum a^2(y_I-y_H)(z_I-z_H) +\sum a^2(y_P-y_H)(z_P-z_H) \overset{(?)}= 0\ . $$ We expand inside the sums, cancel the terms involving $y_Hz_H$, and group by the coefficients of $H$: $$ -\sum a^2 y_Iz_I +\sum a^2 y_Pz_P - \sum(z_P-z_I)(a^2y_H+b^2x_H) \overset{(?)}= 0\ . $$ The first sum is simple, $ S_1 =\sum a^2 y_Iz_I= \sum a^2\; bc/(a+b+c)^2= abc/(a+b+c)^2\sum a= abc/(a+b+c) $.
The second sum can be obtained from the three circle ecuations of $P$, the sum $S_2=\sum a^2 y_Pz_P$ is the quadratic part, then we have a linear part (if the factor $1=\sum x_P$ is omitted) for each of the three equations, considering the matrix of the linears involved, $$ M = M_P = \begin{bmatrix} 0 & \frac{a c^{2}}{a + b} & \frac{a b^{2}}{a + c} \\ \frac{b c^{2}}{a + b} & 0 & \frac{a^{2} b}{b + c} \\ \frac{b^{2} c}{a + c} & \frac{a^{2} c}{b + c} & 0 \end{bmatrix}\ , $$ and we have formally $$ S_2 \begin{bmatrix} 1\\1\\1 \end{bmatrix} = M \begin{bmatrix} x_P\\y_P\\z_P \end{bmatrix} \ , $$ and from here $$ S_2 [1\ 1\ 1] M^{-1} \begin{bmatrix} 1\\1\\1 \end{bmatrix} = [1\ 1\ 1] M^{-1} M \begin{bmatrix} x_P\\y_P\\z_P \end{bmatrix} = [1\ 1\ 1] \begin{bmatrix} x_P\\y_P\\z_P \end{bmatrix} = 1 \ . $$ This makes it humanly rather computable $S_2=\sum a^2 y_PzP$ as $1/S_1=$(sum of the entries in $M^{-1}$).
(And this sum of entries is -up to a rather simple factor- the ugly denominator in the formulas. We expect it also in the next sums.
A third sum is $S_3=\sum x_I(b^2z_H+c^2y_H)=\sum (ab^2+a^2b)/(a+b+c)z_H$. This can be computed using the matrix $M=M_H$ extracted from the linear equations satisfied by definition by $x_H,y_H,z_H$, so we have a relation of the shape $$ M_H \begin{bmatrix} x_H\\y_H\\z_H \end{bmatrix} = \begin{bmatrix} 1\\1\\1 \end{bmatrix} $$ (or some similar one, with an other constant vector). Then we can proceed as above to isolate the needed sum. Unfortunately, the computation has the same degree of complexity as the "straightforward computation".
The final sum $S_4=\sum x_P(b^2z_H+c^2y_H)$ has a double complexity, we need the right combination of the $3\times 3$ linear relations satisfied by $P$, times those satisfied by $H$. But if the "right splitting" for the $M_H$ in $S_3$ was not found, this is useless.
I decided to stop here, still needing the "right splitting", or the right insight.
(3) Some observations:
Connected to the points in the problem, and searching for a synthetic proof, it may be good to also mention the following:
The center 1, incenter, is fundamental by construction, needs no explanation.
How to construct 500 in a "simple(r) manner"? On the line 1,500 (short for $X(1),X(500)$) there are also the following points: 1, 30, 79, 500, 554, 1081, 1464, 1717, 1836, 3058, 3649, 3655, 3656, 3782, 4654, 4854, 5160, 5434, and many others. In particular, one can construct "simpler" the point 500
by drawing a parallel through 1 (incenter) to the Euler line 2 (centroid), 3 (circumcenter), 4 (orthocenter), 5 (9P circle), 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, ... (On this parallel we have some more or less involved points, 1, 30, 79, 500, 554, 1081, 1464, 1717, 1836, 3058, 3649, 3655, 3656, 3782, ...)
and intersecting this parallel with the line through 3 (circumcenter), 6 (Lemoine, symmedian point), 15, 16, 32, 39, 50, 52, 58, 61, 62, 182, 187, 216, 284, 371, 372, 386, 389, 500, 511, ...
because the above two lines are so pregnant, we do not need any further choices like 35, 500, 1154, 1399, 2003, 2594, 6097, 6149, 7186, ... which are rather involved.
How to construct now the last needed point 501 in a "simpler manner"? By definition (Miquel point associated to the incenter / to the incentral triangle), it is the intersection of three circles, but how to express it using (only) lines? First of all, it is on the lines
and may of the above lines are rather involved, but it also lies on the remarkable line below:
The line 500, 501 passes through 2070, which is the inverse in circumcircle of 5.
Note also the mid point 5434 of 1,500, which is the center of the Hatzipolakis circle, with a possible definition (in terms of "simpler points") as 1,30 $\cap$ 3, 81. This point has a geometric property that is simply expresssed in terms of the objects involved in our problem, namely, citing from the ETC, and using our notations:
Let $\Delta DEF$ be the cevian triangle of $I=X(1)$ (the incenter). Let $N_A$ be the nine-point center of triangle $IEF$, and define $N_B$ and $N_C$ cyclically. The points $I$, $N_A$, $N_B$, $N_C$ are concyclic, and their circle [is the circle], described by Antreas Hatzipolakis, April 17, 2013.
Note that the rather complicated point 2605 has also equal distances to 1 and 501.
(4) Some picutures.
Here are some geogebra screen shots, that show what i have tried synthetically.
$I,H,P$, the red Euler line $234$, the parallel to it through $1$, and the points $36,2070$, bird view:
$I,H,P$, the red Euler line $2,3,4$, the parallel to it through $1$, and the points $6,21$, closer view:
Here, one sees my subjectively only chance try to attempt to solve synthetically, the point $6$, the symmedians, and their cross points with circles.
Finally, a closer look in the same geogebra shot:
$I,H,P$, the red Euler line $2,3,4$, the parallel to it through $1$, triangle view.
$I,H,P$, the red Euler line $2,3,4$, the parallel to it through $1$, triangle view.">
One picture from an other geogebra session, here one sees maybe better, that $1,500,501$ are not colinear, the line $1,500$ is parallel to the Euler line, and the line $500,501$ will cross the Euler line $2,3,4$ somewhere (in $2070$) when the screen get a bigger height.
$I,H,P$ are on the same circle, but not colinear">
(5) I will still seek a synthetic proof next days, but this also not with "bare hands", i will let the computer search for the "right inversion", and/or the "right triangle to see the structure inside", and/or the "useful constellation(s) of four points on the same circle"...
Maybe this is a good point to submit the results and thoughts so far.
$I,H,P$, the red Euler line 
$I,H,P$, the red Euler line
This is the kind of problem one is tempted to apply the violence of barycentric coordinates to.
The orthocenter of the incentral triangle is $X(500)$, its trilinear/barycentric coordinates are given by polynomials with degree $6/7$ in $a,b,c$. The trilinear/barycentric coordinates of $I$ are simply $[1;1;1]/[a;b;c]$ and one just needs to find the trilinear/barycentric coordinates of $P$ and let a CAS check that $P+I=2H$. The barycentric coordinates of $P$ can be found by considering that $\frac{BF}{BD}=\frac{b+c}{b+a}$ and that $PD$ is the radical axis of two circles with known centers.
On the other hand I agree that "a CAS can solve this with very little training" is not a satisfactory answer.