Midpoints of Altitudes
by v_Enhance, Aug 20, 2012, 7:44 PM
The summer after this year's MOP, I saw exactly three problems that involve the midpoint of an altitude. (Amusingly, I have not seen any before.) All of them, apparently, can be solved with the same lemma:
Lemma: Let
be the midpoint of the altitude from 
in 
. The incircle touches 
at 
; the -excircle touches at 
. Then , , and 
are collinear, as are , 
, and .
![[asy]
/* DRAGON 0.0.9.6
Homemade Script by v_Enhance. */
import olympiad; import cse5; size(11cm); real lsf=0.4000; real lisf=2011.0; defaultpen(fontsize(10pt)); real xmin=-4.07; real xmax=4.17; real ymin=-2.36; real ymax=2.44;
/* Initialize Objects */
pair A = (-1.5, 2.0);
pair B = (-2.0, 0.0);
pair C = (0.5, 0.0);
pair I = incenter(A,B,C);
pair K = foot(A,B,C);
pair M = midpoint(A--K);
pair D = foot(I,B,C);
pair I_a = (2)*(IntersectionPoint(circumcircle(A,B,C),Line(A,I,lisf),1))-I;
pair X_B = foot(I_a,A,B);
pair X_C = foot(I_a,A,C);
pair D_0 = foot(I_a,B,C);
pair D_1 = (2)*(I)-D;
/* Draw objects */
dot(A);
dot(B);
dot(C);
draw(A--B, rgb(0.6,0.2,0.0));
draw(B--C, rgb(0.6,0.2,0.0));
draw(C--A, rgb(0.6,0.2,0.0));
draw(incircle(A,B,C), rgb(0.6,0.6,0.0) + linewidth(1.2));
dot(I);
dot(K);
draw(A--K, rgb(0.2,1.0,0.0) + dotted);
dot(M);
dot(D);
dot(I_a);
dot(X_B);
dot(X_C);
dot(D_0);
draw(CirclebyPoint(I_a,D_0), rgb(0.6,0.6,0.0) + linewidth(1.2) + dotted);
draw(B--X_B, rgb(0.6,0.6,0.6) + linetype("4 4"));
draw(C--X_C, rgb(0.6,0.6,0.6) + linetype("4 4"));
draw(M--I_a, rgb(0.2,1.0,1.0) + linewidth(1.2) + linetype("8 4 0 4"));
dot(D_1);
draw(A--D_0, rgb(0.2,1.0,0.0) + linewidth(1.0) + linetype("4 4"));
draw(A--I_a, rgb(0.2,1.0,0.0));
draw(M--D_0, rgb(0.6,0.6,0.6));
/* Label points */
label("$A$", A, lsf*dir(45));
label("$B$", B, lsf*dir(45));
label("$C$", C, lsf*dir(45));
label("$I$", I, lsf*dir(45));
label("$K$", K, lsf*dir(45));
label("$A_0$", M, lsf*dir(45));
label("$D$", D, lsf*dir(45));
label("$I_a$", I_a, lsf*dir(45));
label("$X_B$", X_B, lsf*dir(45));
label("$X_C$", X_C, lsf*dir(45));
label("$D_0$", D_0, lsf*dir(45));
label("$D_1$", D_1, lsf*dir(45));
/* Clip the image */
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]](//latex.artofproblemsolving.com/c/1/3/c13d0e9ac3c76ecf2a41aecc41f6492e7ce635f1.png)
This proof was provided to me by someone who is good at geo. To her credit, she both conjectured and proved it while doing one of the aforementoined problems.
====================================================================
Now, given this, here are the three problems, in the order they appeared to me.
By the way,
.
Lemma: Let
be the midpoint of the altitude from 
in 
. The incircle touches 
at 
; the -excircle touches at 
. Then , , and 
are collinear, as are , 
, and .![[asy]
/* DRAGON 0.0.9.6
Homemade Script by v_Enhance. */
import olympiad; import cse5; size(11cm); real lsf=0.4000; real lisf=2011.0; defaultpen(fontsize(10pt)); real xmin=-4.07; real xmax=4.17; real ymin=-2.36; real ymax=2.44;
/* Initialize Objects */
pair A = (-1.5, 2.0);
pair B = (-2.0, 0.0);
pair C = (0.5, 0.0);
pair I = incenter(A,B,C);
pair K = foot(A,B,C);
pair M = midpoint(A--K);
pair D = foot(I,B,C);
pair I_a = (2)*(IntersectionPoint(circumcircle(A,B,C),Line(A,I,lisf),1))-I;
pair X_B = foot(I_a,A,B);
pair X_C = foot(I_a,A,C);
pair D_0 = foot(I_a,B,C);
pair D_1 = (2)*(I)-D;
/* Draw objects */
dot(A);
dot(B);
dot(C);
draw(A--B, rgb(0.6,0.2,0.0));
draw(B--C, rgb(0.6,0.2,0.0));
draw(C--A, rgb(0.6,0.2,0.0));
draw(incircle(A,B,C), rgb(0.6,0.6,0.0) + linewidth(1.2));
dot(I);
dot(K);
draw(A--K, rgb(0.2,1.0,0.0) + dotted);
dot(M);
dot(D);
dot(I_a);
dot(X_B);
dot(X_C);
dot(D_0);
draw(CirclebyPoint(I_a,D_0), rgb(0.6,0.6,0.0) + linewidth(1.2) + dotted);
draw(B--X_B, rgb(0.6,0.6,0.6) + linetype("4 4"));
draw(C--X_C, rgb(0.6,0.6,0.6) + linetype("4 4"));
draw(M--I_a, rgb(0.2,1.0,1.0) + linewidth(1.2) + linetype("8 4 0 4"));
dot(D_1);
draw(A--D_0, rgb(0.2,1.0,0.0) + linewidth(1.0) + linetype("4 4"));
draw(A--I_a, rgb(0.2,1.0,0.0));
draw(M--D_0, rgb(0.6,0.6,0.6));
/* Label points */
label("$A$", A, lsf*dir(45));
label("$B$", B, lsf*dir(45));
label("$C$", C, lsf*dir(45));
label("$I$", I, lsf*dir(45));
label("$K$", K, lsf*dir(45));
label("$A_0$", M, lsf*dir(45));
label("$D$", D, lsf*dir(45));
label("$I_a$", I_a, lsf*dir(45));
label("$X_B$", X_B, lsf*dir(45));
label("$X_C$", X_C, lsf*dir(45));
label("$D_0$", D_0, lsf*dir(45));
label("$D_1$", D_1, lsf*dir(45));
/* Clip the image */
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]](http://latex.artofproblemsolving.com/c/1/3/c13d0e9ac3c76ecf2a41aecc41f6492e7ce635f1.png)
This proof was provided to me by someone who is good at geo. To her credit, she both conjectured and proved it while doing one of the aforementoined problems.
be the point on the incircle diametrically opposite to 
and 
are parallel with midpoints 
is perpendicular to 
and 
.
and 
are similar, and so by manipulating the similarity ratios you get 
.
.
====================================================================
Now, given this, here are the three problems, in the order they appeared to me.
- (Vietnam TST 2003, Problem 2) Given a triangle
. Let 
be the circumcenter of this triangle . Let 
, 
, 
be the feet of the altitudes of triangle from the vertices , 
, 
, respectively. Denote by 
, 
, 
the midpoints of these altitudes 
, 
, 
, respectively. The incircle of triangle has center and touches the sides , 
, 
at the points , 
, 
, respectively. Prove that the four lines 
, 
, 
and 
are concurrent. (When the point concides with , we consider the line as an arbitrary line passing through .
- (Iran TST 2009, Problem 9) In triangle , , and are the points of tangency of incircle with the center of to , and respectively. Let
be the foot of the perpendicular from to 
. 
is on 
such that 
. If is the orthocenter of 
, prove that 
bisects .
- (ISL 2002, G7) The incircle
of the acute-angled triangle is tangent to its side at a point . Let 
be an altitude of triangle , and let be the midpoint of the segment . If 
is the common point of the circle and the line 
(distinct from ), then prove that the incircle and the circumcircle of triangle 
are tangent to each other at the point .
.
be the circumcenter of this triangle 
,
,
be the feet of the altitudes of triangle 
,
,
,
,
the midpoints of these altitudes 
,
,
,
,
at the points 
,
,
,
,
and 
are concurrent. (When the point 
be the foot of the perpendicular from 
.
is on 
such that 
.
,
bisects 
of the acute-angled triangle 
be an altitude of triangle 
is the common point of the circle 
(distinct from 
are tangent to each other at the point By the way,
.This post has been edited 1 time. Last edited by v_Enhance, Aug 20, 2012, 7:48 PM
![\[ \det \left[ \begin{array}{ccc} 2a^2 & a^2+c^2-b^2 & a^2+b^2-c^2 \\ 0 & a+c-b & a+b-c \\ -a & b & c \end{array} \right] = 0 \]](http://latex.artofproblemsolving.com/2/4/3/24325b4a27bca6e7038cdcd690f3d2660018d72b.png)
I didn't feel like expanding that at time of writing.
and 
are both nicer than 
,
be a diameter of the incircle, and extend 
to meet 
.
,
.
is harmonic.
or 
,![\[(CK-BK)r\tan\angle DKN=BK\cdot CK\rightarrow\tan\angle DKN=\frac{BK\cdot CK}{r(CK-BK)}\]](http://latex.artofproblemsolving.com/f/4/8/f4850c50238f0a160082af6e8bcfab96b5701c15.png)
(using the lemma), and 
(letting the parallel from 
and using PoP).
,
by a lemma of harmonic divisions 
.
intersect 
at 
and the incircle is tangent to 
are collinear with 
April 2023
February 2023