A Simple Lemma Revisited
by liberator, Aug 15, 2014, 9:52 PM
Several questions in olympiads are very closely linked, and I present one example where the diagram bears resemblance to that of A Simple Lemma.
Problem 1: In an acute scalene triangle
the bisector of the acute angle between the altitudes 
and 
meets the sides 
and 
at 
and 
respectively. The bisector of the angle 
intersects the segment joining the orthocenter of and the midpoint of at point 
. Prove that 
is cyclic.
[All-Russian Olympiad 2000 Grade 10 Q3]
Problem 2: Let be an acute-angled triangle with 
. Let 
be the orthocenter of triangle , and let 
be the midpoint of the side . Let 
be points on 
respectively such that 
and the points , , are collinear. Prove that the line 
is perpendicular to the common chord of the circumscribed circles of 
and 
.
[ISL 2005 G5]
Problem 3: Let be an acute triangle with orthocenter . The external bisector of 
intersects 
and 
at and respectively.The internal bisector of 
intersects the circumcircle of again at . Show that 
passes through the midpoint of .
[British FST-1 2014 Q3, Vietnam TST 2006 Q1]
Visualization:
![[asy]
/* Free script by liberator, 15 August 2014 */
unitsize(3cm);
defaultpen(fontsize(10pt));
pointpen=black;
/* Initialize objects */
pair A = dir(110);
pair B = dir(200);
pair C = dir(340);
pair Ap = dir(-A);
pair H = orthocenter(A,B,C);
pair Y = foot(B,C,A);
pair Z = foot(C,A,B);
pair D1 = bisectorpoint(Z,H,B);
pair D = intersectionpoint(A--B, H--D1);
pair E1 = bisectorpoint(C,H,Y);
pair E = intersectionpoint(C--A, H--E1);
pair K = intersectionpoint(H--Ap, circumcircle(A,D,E));
pair P = foot(A,K,H);
pair M = midpoint(B--C);
/* Draw objects */
draw(A--B--C--cycle, rgb(0.4,0.6,0.8)+linewidth(1));
draw(H--Z--Y--cycle, rgb(0.4,0.6,0.8));
draw(D--E, rgb(0.4,0.6,0.8)+linewidth(1));
draw(A--H, rgb(0.4,0.6,0.8));
draw(B--H, rgb(0.4,0.6,0.8));
draw(C--H, rgb(0.4,0.6,0.8));
draw(P--Ap, rgb(0.4,0.6,0.8)+linewidth(1)+dashed);
draw(unitcircle, rgb(0,0.6,0));
draw(circumcircle(A,D,E), rgb(0.9,0,0));
draw(circumcircle(A,Z,Y), rgb(0.9,0,0));
/* Place dots on and label each point */
dot(A); label("$A$", A, dir(A));
dot(B); label("$B$", B, dir(B));
dot(C); label("$C$", C, dir(C));
dot(Ap); label("$A'$", Ap, dir(Ap));
dot(D); label("$D$", D, dir(D));
dot(E); label("$E$", E, dir(E));
dot(H); label("$H$", H, 1.2*dir(-100));
dot(M); label("$M$", M, dir(M));
dot(K); label("$K (K')$", K, dir(-70));
dot(Y); label("$Y$", Y, dir(0));
dot(Z); label("$Z$", Z, dir(Z));
dot(P); label("$P$", P, dir(P));
[/asy]](//latex.artofproblemsolving.com/3/6/f/36f114c89e48e31691bbe9c0d778df238c78e51e.png)
As you may have noticed, I have adapted some of the questions so as to keep notation consistent. In this way, it is easier to see the similarities between solutions and enables a better understanding of how the problems are linked. Of course, we could use roughly the same solution for each question, but I have presented three different (synthetic) proofs, so that the reader may try as an exercise to adapt each different proof to the other questions.
Problem 1 solution: Let
be antipodal to on the 
, and the orthocenter. Clearly, 
and 
are symmetrical about the angle bisector 
, and 
, since 
is a parallelogram from A Simple Lemma.
Since
, we get 
, so 
. But 
, so as 
is angle bisector, hence
![\[\frac {DZ}{CD} = \frac {EZ}{EB} = \frac {AY}{AB} = \frac {HK}{KA'}.\]](//latex.artofproblemsolving.com/d/a/e/dae1fe0d43b398df7c3f1a0d51f6243beee853a4.png)
It follows that
and 
, hence is cyclic.
Problem 2 solution: Denote
the Miquel point of 
. By Brochard, 
, so it suffices to show that 
, where 
.
Let
be the radical center of 
, and let 
. Hence 
. By Maclaurin's theorem, 
. By considering the power of 
w.r.t 
, we have ![\[RD \cdot RE = RH \cdot RN = RP' \cdot RA,\]](//latex.artofproblemsolving.com/1/b/c/1bc8a4a1bd48cd2d0c6e4b76cc8ea4e2a047fba6.png)
so that 
.
Problem 3 solution: Let
and be the antipode of w.r.t . By A Simple Lemma, 
are collinear.
Now note that
. Since 
we have 
which means 
, hence 
is cyclic, so 
and the result follows.
For more, you can see http://jl.ayme.pagesperso-orange.fr/Docs/un%20leitmotif.pdf.
Problem 1: In an acute scalene triangle
the bisector of the acute angle between the altitudes 
and 
meets the sides 
and 
at 
and 
respectively. The bisector of the angle 
intersects the segment joining the orthocenter of and the midpoint of at point 
. Prove that 
is cyclic.[All-Russian Olympiad 2000 Grade 10 Q3]
Problem 2: Let
be an acute-angled triangle with 
. Let 
be the orthocenter of triangle , and let 
be the midpoint of the side . Let 
be points on 
respectively such that 
and the points , , are collinear. Prove that the line 
is perpendicular to the common chord of the circumscribed circles of 
and 
.[ISL 2005 G5]
Problem 3: Let
be an acute triangle with orthocenter . The external bisector of 
intersects 
and 
at and respectively.The internal bisector of 
intersects the circumcircle of again at . Show that 
passes through the midpoint of .[British FST-1 2014 Q3, Vietnam TST 2006 Q1]
Visualization:
![[asy]
/* Free script by liberator, 15 August 2014 */
unitsize(3cm);
defaultpen(fontsize(10pt));
pointpen=black;
/* Initialize objects */
pair A = dir(110);
pair B = dir(200);
pair C = dir(340);
pair Ap = dir(-A);
pair H = orthocenter(A,B,C);
pair Y = foot(B,C,A);
pair Z = foot(C,A,B);
pair D1 = bisectorpoint(Z,H,B);
pair D = intersectionpoint(A--B, H--D1);
pair E1 = bisectorpoint(C,H,Y);
pair E = intersectionpoint(C--A, H--E1);
pair K = intersectionpoint(H--Ap, circumcircle(A,D,E));
pair P = foot(A,K,H);
pair M = midpoint(B--C);
/* Draw objects */
draw(A--B--C--cycle, rgb(0.4,0.6,0.8)+linewidth(1));
draw(H--Z--Y--cycle, rgb(0.4,0.6,0.8));
draw(D--E, rgb(0.4,0.6,0.8)+linewidth(1));
draw(A--H, rgb(0.4,0.6,0.8));
draw(B--H, rgb(0.4,0.6,0.8));
draw(C--H, rgb(0.4,0.6,0.8));
draw(P--Ap, rgb(0.4,0.6,0.8)+linewidth(1)+dashed);
draw(unitcircle, rgb(0,0.6,0));
draw(circumcircle(A,D,E), rgb(0.9,0,0));
draw(circumcircle(A,Z,Y), rgb(0.9,0,0));
/* Place dots on and label each point */
dot(A); label("$A$", A, dir(A));
dot(B); label("$B$", B, dir(B));
dot(C); label("$C$", C, dir(C));
dot(Ap); label("$A'$", Ap, dir(Ap));
dot(D); label("$D$", D, dir(D));
dot(E); label("$E$", E, dir(E));
dot(H); label("$H$", H, 1.2*dir(-100));
dot(M); label("$M$", M, dir(M));
dot(K); label("$K (K')$", K, dir(-70));
dot(Y); label("$Y$", Y, dir(0));
dot(Z); label("$Z$", Z, dir(Z));
dot(P); label("$P$", P, dir(P));
[/asy]](http://latex.artofproblemsolving.com/3/6/f/36f114c89e48e31691bbe9c0d778df238c78e51e.png)
As you may have noticed, I have adapted some of the questions so as to keep notation consistent. In this way, it is easier to see the similarities between solutions and enables a better understanding of how the problems are linked. Of course, we could use roughly the same solution for each question, but I have presented three different (synthetic) proofs, so that the reader may try as an exercise to adapt each different proof to the other questions.
Problem 1 solution: Let
be antipodal to on the 
, and the orthocenter. Clearly, 
and 
are symmetrical about the angle bisector 
, and 
, since 
is a parallelogram from A Simple Lemma.Since
, we get 
, so 
. But 
, so as 
is angle bisector, hence![\[\frac {DZ}{CD} = \frac {EZ}{EB} = \frac {AY}{AB} = \frac {HK}{KA'}.\]](http://latex.artofproblemsolving.com/d/a/e/dae1fe0d43b398df7c3f1a0d51f6243beee853a4.png)
It follows that
and 
, hence is cyclic.Problem 2 solution: Denote
the Miquel point of 
. By Brochard, 
, so it suffices to show that 
, where 
.Let
be the radical center of 
, and let 
. Hence 
. By Maclaurin's theorem, 
. By considering the power of 
w.r.t 
, we have ![\[RD \cdot RE = RH \cdot RN = RP' \cdot RA,\]](http://latex.artofproblemsolving.com/1/b/c/1bc8a4a1bd48cd2d0c6e4b76cc8ea4e2a047fba6.png)
so that 
.Problem 3 solution: Let
and be the antipode of w.r.t . By A Simple Lemma, 
are collinear.Now note that
. Since 
we have 
which means 
, hence 
is cyclic, so 
and the result follows.For more, you can see http://jl.ayme.pagesperso-orange.fr/Docs/un%20leitmotif.pdf.
This post has been edited 9 times. Last edited by liberator, Nov 2, 2014, 11:59 AM
so 
bisects 
and then easily 
.
October 2018
April 2018