geometry problem

[stephanson12]

Let O be the center of an acute-angled triangle ABC. OA cuts the altitudes from B and
C at points P and Q, respectively. Let H be the orthocenter and let M be the midpoint of [BC]. Prove
that the center of (HPQ) is at AM.

All hints are welcome! (Sorry for my bad english)

 

[The Highlander]

Let O be the center of an acute-angled triangle ABC. OA cuts the altitudes from B and
C at points P and Q, respectively. Let H be the orthocenter and let M be the midpoint of [BC]. Prove
that the center of (HPQ) is at AM.

All hints are welcome! (Sorry for my bad english)

Hi @stephanson12

By "center" do you mean the centroid?

(And did you mean to write "the center of (ΔHPQ) lies on the line AM"?)

Please provide a sketch of your interpretation of the problem (to illustrate what you have tried) and, if any further advice is required after you have done that, someone will provide further comment, I'm sure.

I would suggest you should also have a look at this website (first?).

Hope that helps.

 

[stephanson12]

circumcenter

 

[Aion]

I tried drawing it in GeoGebra. Maybe I did something wrong, but it doesn't seem like the circumcenter $O'$' of $\Delta HPQ$ lies on the line $AM$.

 

[pka]

Let O be the center of an acute-angled triangle ABC. OA cuts the altitudes from B and
C at points P and Q, respectively. Let H be the orthocenter and let M be the midpoint of [BC]. Prove
that the center of (HPQ) is at AM.

It seems that there may be a language problem here. Three terms are possible: circumcenter, incenter, or orthocenter.
You specify circumcenter. If $\Delta{ABC}$ is an acute triangle then its circumcenter, $\mathrm{O},$ is the point that is the intersection of the perpendicular bisectors of two sides of triangle, As such $m(\overline{OA})=m(\overline{OB})=m(\overline{OC})$.
That is because $\mathrm{O},$ is the center of the circle which includes all three vertices of the triangle.
The altitudes of a triangle also intersect at a common point called the orthocenter.

 

[Aion]

I think I made a mistake yesterday. I tried it again and got the following.


Where $O$ is the circumcenter of $\Delta ABC$, $H$ is the orthocenter of $\Delta ABC$,$O'$ is the circumcenter of $\Delta HPQ$, and $M$ is the midpoint of segment $BC$. From this, it appears to be a true statement. Now you have to prove it I guess.

 

[Dr.Peterson]

Let O be the circumcenter of an acute-angled triangle ABC. OA cuts the altitudes from B and
C at points P and Q, respectively. Let H be the orthocenter and let M be the midpoint of [BC]. Prove
that the circumcenter of (HPQ) is on AM.

All hints are welcome! (Sorry for my bad english)

I've added "circum" where needed and changed "at" to "on" (the only error in your English).

Here's my version of the figure, with more labels:

​

Now, how might we prove it?

Have you tried anything yet? We'd like you to tell us that. I don't yet see any hints other than what I've done: make a picture. (But I do see lots of parallel lines, similar triangles, and the like.)

Also, is this from a class that might imply what sort of methods to consider, or from a contest, or what?

 

[Aion]

I'm not particularly good at geometry but I think I've devised one approach to the problem. Without loss of generality let the unit circle of triangle $ABC$ be the circumcircle. Let $A=-i$, $B=a+bi$ and $C=c+di$.






First, let's consider the equation of line through points. Suppose $Z$ is an arbitrary point on the line through $W_1$ and $W_2$. Since the vector from $W_1$ to$Z$ is a multiple of the vector from $W_1$ to $W_2$, in terms of complex numbers we have $Z-W_1=\lambda(W_2-W_1)$ for some real $\lambda$. Now $\lambda=\bar{\lambda}$ hence

$$\frac{Z-W_1}{W_2-W_1}=\frac{\bar{Z}-\bar{W_1}}{\bar{W_2-W_1}}$$and any $Z$ satisfying the above equation corresponds to a point on the line through $W_1$ and $W_2$. Next, consider the line passing through a point $C$ and perpendicular to the line through $W_1$ and$W_2$. Let $Z$ be on this line. Then the vector from $C$ to $Z$ is perpendicular to the vector from $W_1$ to $W_2$. In terms of complex numbers, we get $Z-C=i\lambda(W_2-W_1)$ for some real $\lambda$.

Hence $$\frac{Z-C}{i(W_2-W_1)}=\frac{\bar{Z}-\bar{C}}{-i(\bar{W_2}-\bar{W_1})}$$.

In the case when $W_1$ and $W_2$ are on the unit circle we have $W_1\bar{W_1}=W_2\bar{W_2}=1$. Multiplying the numerators and denominators of the right side of the two equations above by $W_1W_2$ we can simplify them as

$$Z+W_1W_2\bar{Z}=W_1+W_2$$ and $$Z-W_1W_2\bar{Z}=C-W_1W_2\bar{C}$$
Now since $A$, $B$and $C$ are points on the unit circle these formulas can be applied immediately to compute $Q$ and $P$.

The line through $C$ perpendicular to $AB$ is given by

$$Z-AB\bar{Z}=C-AB\bar{C}$$
And the line through $0$ and $-i$ is $Z=-\bar{Z}$. Since these two lines intersect at $Q$, solving this system of equations for $Z$ gives us $Q$. Hence

$$Q=\frac{C-AB\bar{C}}{1+AB}$$
Similarly, the line through $B$ perpendicular to $AC$ is

$$Z-AC\bar{Z}=B-AC\bar{B}$$
And this line intersects $Z=-\bar{Z}$ at $P$. Hence

$$P=\frac{B-AC\bar{B}}{1+AC}$$
Now with complex numbers, the orthocenter $H=A+B+C$. Hence each vertices of the triangle $HPQ$ can be computed and a similar method could now be used to determine $O'$. Now $M =\frac{B+C}{2}$and we must show that $M, O'$ and $A$ are colinear. In other words that

$$\frac{A-O'}{O'-M}=\lambda\in \mathbb{R}$$

 

[Aion]

I know the OP hasn't responded and this is considered a dead thread. But I've been thinking about this problem for a while and don't think complex numbers is the right approach since solving for the intersection of lines becomes quite complicated. Here's my approach using the above model anyway.

First, let's scale the vertices of the triangle $HPQ$ by 2, so let $Q'=2Q$, $P'=2P$ and $H'=2H$. Scaling by a constant factor preserves angles and ratio of distances hence the circumcenter of this triangle is scaled by the same factor. So if $O'$ is the circumcenter of the scaled triangle $H'P'Q'$ then $O'=2O$ where $O$ is the circumcenter of the original triangle $HPQ.$



Let $E$ be the midpoint of segment $Q'P'$ and $D$ be the midpoint of segment $P'H'$ such that $E=Q+P$ and $D=P+H$. Let $u$ be the vector from $D$ to $H'$and $v$ be the vector from $E$ to $P'$, then $u=H-P$ and $v=P-Q$. Then the vector perpendicular to $u$ is $iu$ and the vector perpendicular to $v$ is $iv$. Consider the line passing through $D$ perpendicular to $u$. If $Z$ is any point on this line then it must satisfy both of the following equations.

$$Z=D+itu=P+H+it(H-P)$$$$\bar{Z}=\bar{D}+\bar{i}\bar{t}\bar{u}=\bar{P}+\bar{H}-i\bar{t}(\bar{H}-\bar{P})$$
Since $t=\bar{t}$, the locus of the midpoint normal through $P'H'$is given by

$$\frac{Z-D}{iu}=\frac{\bar{Z}-\bar{D}}{\bar{i}\bar{u}}$$
Simplifying we obtain
$$(Z-D)\bar{u}+(\bar{Z}-\bar{D})u=0$$
or
$$(Z-(H+P))(\bar{H}-\bar{P})+(\bar{Z}-(\bar{H}+\bar{P}))(H-P)=0$$
By similar reasoning, or by symmetry we can conclude that the locus of the midpoint normal through $Q'P'$ is determined by

$$(Z-E)\bar{v}+(\bar{Z}-\bar{E})v=0$$or
$$(Z-(P+Q))(\bar{P}-\bar{Q})+(\bar{Z}-(\bar{P}+\bar{Q}))(P-Q)=0$$
On solving these two equations in terms of $Z$ we should obtain the coordinate of $O'$ (if I've calculated this correctly so far). As you can see the solution would be quite complicated using this method.

 

[Aion]

From

$$(Z-D)\bar{u}+(\bar{Z}-\bar{D})u=0$$$$(Z-E)\bar{v}+(\bar{Z}-\bar{E})v=0$$
First, solve for $\bar{Z}$ in the first equation. We get
$$\bar{Z}=\frac{(D-Z)\bar{u}}{u}+\bar{D}$$
Then substitute this into the second equation. We have

$$(Z-E)\bar{v}+ \left(\frac{(D-Z)\bar{u}}{u}+D-E\right)v=0$$
Set $v'=\frac{\bar{v}}{v}$ and $u'=\frac{\bar{u}}{u}$, then solve for $Z$

$$Z=\frac{E(v'+1)-D(u'+1)}{v'-u'}$$$$Z=O'=\frac{(Q+P)(\frac{\bar{v}}{v}+1)-(P+H)(\frac{\bar{u}}{u}+1)}{\frac{\bar{v}}{v}-\frac{\bar{u}}{u}}=\frac{(P+Q)(\frac{\bar{P}-\bar{Q}}{P-Q}+1)-(H+P)(\frac{\bar{H}-\bar{P}}{H-P}+1)}{\frac{\bar{P}-\bar{Q}}{P-Q}-\frac{\bar{H}-\bar{P}}{H-P}}$$

 

[Aion]

$$Z=\frac{Ev'+\bar{E}-D(u'+1)}{v'-u'}$$From

$$(Z-D)\bar{u}+(\bar{Z}-\bar{D})u=0$$$$(Z-E)\bar{v}+(\bar{Z}-\bar{E})v=0$$
First, solve for $\bar{Z}$ in the first equation. We get
$$\bar{Z}=\frac{(D-Z)\bar{u}}{u}+\bar{D}$$
Then substitute this into the second equation. We have

$$(Z-E)\bar{v}+ \left(\frac{(D-Z)\bar{u}}{u}+D-E\right)v=0$$
Set $v'=\frac{\bar{v}}{v}$ and $u'=\frac{\bar{u}}{u}$, then solve for $Z$

$$Z=\frac{E(v'+1)-D(u'+1)}{v'-u'}$$$$Z=O'=\frac{(Q+P)(\frac{\bar{v}}{v}+1)-(P+H)(\frac{\bar{u}}{u}+1)}{\frac{\bar{v}}{v}-\frac{\bar{u}}{u}}=\frac{(P+Q)(\frac{\bar{P}-\bar{Q}}{P-Q}+1)-(H+P)(\frac{\bar{H}-\bar{P}}{H-P}+1)}{\frac{\bar{P}-\bar{Q}}{P-Q}-\frac{\bar{H}-\bar{P}}{H-P}}$$

I made an obvious error yesterday,

It should be
$$(Z-E)\bar{v}+\left(\frac{(D-Z)\bar{u}}{u}+D-\bar{E})\right)v=0$$
$$(Z-E)v'+(D-Z)u'+D-\bar{E}=0$$$$Zv'-Ev'+Du'-Zu'+D-\bar{E}=0$$$$Z(v'-u')=Ev'+\bar{E}-D(u'+1)$$$$Z=\frac{Ev'+\bar{E}-D(u'+1)}{v'-u'}$$
It's an even uglier solution but I guess that's the truth.

 

[Aion]

$v'=\frac{\bar{v}}{v}$I made an obvious error yesterday,

It should be
$$(Z-E)\bar{v}+\left(\frac{(D-Z)\bar{u}}{u}+D-\bar{E})\right)v=0$$
$$(Z-E)v'+(D-Z)u'+D-\bar{E}=0$$$$Zv'-Ev'+Du'-Zu'+D-\bar{E}=0$$$$Z(v'-u')=Ev'+\bar{E}-D(u'+1)$$$$Z=\frac{Ev'+\bar{E}-D(u'+1)}{v'-u'}$$
It's an even uglier solution but I guess that's the truth.

Man, I'm sloppy today as well. I will redo everything again.

$$(Z-D)\bar{u}+(\bar{Z}-\bar{D})u=0$$$$(Z-E)\bar{v}+(\bar{Z}-\bar{E})v=0$$
Solve for $\bar{Z}$ in the first equation.

$$\bar{Z}=\frac{(D-Z)\bar{u}}{u}+\bar{D}$$And substitute into the second equation

$$(Z-E)\bar{v}+\left(\frac{(D-Z)\bar{u}}{u}+\bar{D}-\bar{E})\right)v=0$$
Set $v'=\frac{\bar{v}}{v}$ and $u'=\frac{\bar{u}}{u}$, then divide by $v$ in the above expression to obtain

$$(Z-E)v'+(D-Z)u'+\bar{D}-\bar{E}=0$$
Solve for $Z$

$$Z(v'-u')=Ev'+\bar{E}-(Du'+\bar{D})$$$$Z=\frac{Ev'+\bar{E}-(Du'+\bar{D})}{(v'-u')}$$