8

$\triangle ABC$ has altitudes $AD$, $BE$, $CF$. The reflections of $E$, $F$ in $H$ are $E'$, $F'$. The circle $DE'F'$ intersects $BE$, $CF$ at $X$, $Y$. Prove that $XY$ goes through the midpoints of $AB$, $AC$.

I can show that $XY$ is parallel to $BC$ by simply angle-chasing. $EYFX$ is cyclic as well as $APFY$. enter image description here

I also tried showing that $AP$=$PH+HD$

Plato
  • 2,332
  • This is a nice hard problem, it is a very good experience e.g. when training or when searching adequate material for training for a contest, such as a higher level olympiad for a lower level of mathematical knowledge. Please, which is the source (and the level as it was set by the source)? It is really a pitty that the question did not get the deserved attention from the community. I put a plus one in there... – dan_fulea Feb 09 '22 at 04:03

2 Answers2

0

To illustrate the idea of the solution, consider the following picture first, where the mid points $B_1$, $C_1$ of $CA$, $AB$, and the reflection $D'$ of $D$ in $H$ are also present.

math stackexchange question 2598445

We want to show that the points $X,H,D,B_1$ are on a cycle. This would be enough to conclude, since from here we are allowed to write the first equality in the following chain: $$ \widehat{XB_1D} = 180^\circ - \widehat{XHD} = \widehat{XHA} = \widehat{EHA} = \widehat{EFA} = \hat C= \widehat{B_1DC} \ . $$ This gives $XB_1\|DC=BC$. Similarly $YC_1\|BC$. And now combine this with $XY\|BC$ obtained by the pink angle chasing from the picture, to obtain the collinearity of $B_1,C_1,X,Y$.

So let us show that $(XHDB_1)$ is cyclic by considering the two marked angles in $X$ and $B_1$ against the (a posteriori insured) arc $\overset\frown{HD}$. The angle in $X$ is complicated, but we will move it to a simple place via: $$ \hat X = \widehat{E'XD} = \widehat{E'F'D} = \widehat{EFD'} \ . $$ So we need to clear $$\widehat{EFD'} \overset ?= \widehat{HB_1D}\ . $$ After adding the same angle, $\widehat{EFD} = 2\widehat{EFH}=2(90^\circ-\hat C)=\widehat{DB_1C}$, on both sides, we have to clear equivalently: $$ \widehat{D'FD} \overset ?= \widehat{HB_1C}\ . $$ And indeed, let us show the similarity of triangles $\Delta D'FD\sim\Delta HB_1C$. First of all, we clear an angle, $$ \widehat{FDD'} = \widehat{FDA} = \widehat{FCA} = \widehat{HCB_1}\ . $$ It remains to get a proportion, and this is: $$ \frac{D'D}{HC} = 2\cdot \frac{HD}{HC} = 2\cos B =2\cdot\frac{BD}{BA} =2\cdot\frac{DF}{AC} % \text{ from }\Delta BDF\sim\Delta BAC = \frac{DF}{AC/2} = \frac{DF}{CB_1}\ . $$ $\square$

dan_fulea
  • 32,856
-2

Take point $C$ as the origin of a Cartesian coordinate system. Then let $|PD|=a$. It suffices to note that since the line $XPY$ is parallel to the base of the triangle, its equation is $y=a$. Now since the midpoints of $AC$ and $AB$ have coordinates $(x_1,a)$ and $(x_2,a)$ respectively, it's clear that they lie on the line of interest. $\square$

Allawonder
  • 13,327
  • Could someone explain the error? – Allawonder Jan 11 '18 at 05:19
  • How do you know that the midpoints of $AC$ and $AB$ have coordinates $(x1,a)$ and $(x2,a)$? – Intelligenti pauca Jan 11 '18 at 14:20
  • We know the length of the segments, so we know their midpoints. This should be clear. I only omitted stating the x-coordinates explicitly because they eventually don't matter. – Allawonder Jan 11 '18 at 16:35
  • Of course, bot how do you know that the $y$ coordinate is $a$, the same as that of $P$? – Intelligenti pauca Jan 11 '18 at 16:52
  • @Aretino Say half of $|AC|$ is $\alpha$ and half of $|AB|$ is $\beta$ then circles with radii $\alpha$ and $\beta$ and centered at $B$ and $C$ respectively will interest the lines through $AC$ and $AB$ at their midpoints. The positive solutions of the equations connecting these circles and the lines through the sides are the coordinates sought. – Allawonder Jan 16 '18 at 15:12