No, I'm quite sure you can invoke that without proof. But, are you sure that they always are the altitudes? I've drawn a diagram, and I'm pretty sure that this is not necessarily true...

Yeah that's acceptable.

Ah, thank you guys very much!

I looked at it and I think you can use an induction-like method and reduce it to the altitudes.

probability1.01 wrote: I looked at it and I think you can use an induction-like method and reduce it to the altitudes. Okay. All I did was draw a rough diagram and see that it wasn't necessarily the altitudes...

D, E, and F are all midpoints, so DE, EF, and DF must be parallel to their respective bases. If the line through A is perpendicular to DE, then it must also be perpendicular to AB. Same reason with the others, so you have 3 altitudes of ABC (if you extend the lines to the sides of ABC). Or am I doing something wrong?

They aren't midpoints. They lie on the perpendicular bisectors. However, I think you can sort of "induct" by starting them out at the midpoints.

So it wasn't as easy as I thought. Ah well, back to the scratch paper!

Here's a solution: Let $G$, $H$, $I$ be the feet of the perpendiculars from $A,B,C$ to $EF,FD,DE$ respectively. Let $O$ be the circumcenter of $ABC$. Using the two right angles, we have that $\angle ACI=\angle OFD$ (as directed angles). Similarly, $\angle ICB=\angle FDO$, $\angle CBH=\angle ODE$, $\angle HBA=\angle DEO$, $\angle BAG=\angle OEF$, and $\angle GAC=\angle EFO$. By Trig Ceva in $DEF$, $\frac{\sin\angle OFD\sin\angle ODE\sin\angle OEF}{\sin\angle EFO\sin\angle FDO\sin\angle DEO}=1=\frac{\sin\angle ACI\sin\angle CBH\sin\angle BAG}{\sin\angle GAC\sin\angle ICB\sin\angle HBA}$ $1=\frac{\sin\angle ACI\sin\angle CBH\sin\angle BAG}{\sin\angle ICB\sin\angle HBA\sin\angle GAC}.$ By Trig Ceva in $ABC$, this last equation tells us that $AG$, $BH$, $CI$ are concurrent, and we are done. I was a freshman in 1997, and a first-time USAMO qualifier. I didn't know enough geometry to handle this question, but I did solve the other one using Power of a Point.

This problem can be proved using the fact which I mentioned in http://www.artofproblemsolving.com/Forum/topic-12761.html the diagonals of a convex quadrilateral are perpendicular to each other if and only if the sums of squares of the correponding opposite sides are equal.

I found a pretty easy solution...but I'm not sure if its right...it looked TOO easy, in fact... Solution: Since D,E,F lie on the perpendicular bisectors, then they must be the centers of three circles passing through two of A,B,C. Each pair of circles also intersect at one of D,E,F. Also, note that DE,EF, and FD are the lines joining the centers of two circles. But any line perpendicualr to the line joining the center of two circles and also passing through one of their intersection points must be the radical axis of the two circles. It is then well known that the radical axes of three pairwise intersecting circles are concurrent. QED

You can make use of the fact that the perpendiculars in triangle $DEF$ to the sides at pts $M,N,P$ on $DE,EF,DF$ are concurrent iff $DM^{2}-ME^{2}+EN^{2}-NP^{2}+FP^{2}-PD^{2}=0$. Then you just use Pythagorean theorem two times to get this result.

Quote: I found a pretty easy solution...but I'm not sure if its right...it looked TOO easy, in fact... Solution: Since D,E,F lie on the perpendicular bisectors, then they must be the centers of three circles passing through two of A,B,C. Each pair of circles also intersect at one of D,E,F. Also, note that DE,EF, and FD are the lines joining the centers of two circles. But any line perpendicualr to the line joining the center of two circles and also passing through one of their intersection points must be the radical axis of the two circles. It is then well known that the radical axes of three pairwise intersecting circles are concurrent. QED This is correct (and a great solution!), and follows from the construction of the circumcenter of a triangle with a ruler and compass- it's basically the same process of having the radical axes intersect, with the centers of the circles being the vertices of the given triangle. Then, as an extension you know that the point of concurrency is also the circumcenter of triangle DEF when the three circles happen to have the same radius.

Sorry to be reviving an old topic. But I think this question is basically a one liner(probably I may be wrong). Consider triangle $ ABC$ with points $ D,E,F$ on the perpendicular bisectors of $ BC,CA,AB$ respectively. Let $ D \in l , E\in m, F\in n$ respectively. We know that $ l,m,n$ are concurrent at the circumcenter $ O$ of triangle $ ABC$. Now consider the perpendiculars dropped from $ A$ to $ EF$, $ B$ to $ DF$, $ C$ to $ DE$ as $ p,q,r$ respectively. Orthologic Triangles Theorem : Given two triangles $ ABC, A'B'C'$ respectively. if the perpendiculars from $ A$ to $ B'C'$, $ B$ to $ C'A'$, $ C$ to $ A'B'$ are concurrent if and only if the perpendiculars from $ A'$ to $ BC$, $ B'$ to $ CA$, $ C'$ to $ AB$ respectively are concurrent. (Darij has written about it many times in geometry forums) So here, since $ l,m,n$ are concurrent, it implies obviously that $ p,q,r$ are concurrent . (since $ l$ perpendicular to $ BC$, etc). $ \text{QED}$.

Suppose that the feet of the perpendiculars from $A$ and $P$ to $EH$ are $H$ and $K$ respectively. Then \begin{align*}AF^2-AE^2&=(AH^2+FH^2)-(AH^2+EH^2)\\&=FH^2-EH^2\\&=(FH+EH)(FH-EH)\\&=FE(FH-EH).\end{align*} Similarly $PF^2-PE^2=FE(FK-EK)$. So $H$ and $K$ coincide iff $AF^2-AE^2=PF^2-PE^2$, i.e. $P$ lies on the line through $A$ perpendicular to $EF$ iff $PF^2-PE^2=AF^2-AE^2$. Thus if $P$ is the intersection of the line through $A$ perpendicular to $EF$ and the line through $B$ perpendicular to $FD$, then $PF^2-PE^2=AF^2-AE^2$ and $PD^2-PF^2=BD^2-BF^2$. Hence \[PD^2-PE^2=AF^2-BF^2+BD^2-AE^2.\] But note that $F$ is equidistant from $A$ and $B$, so $AF^2=BF^2$. Similarly we have $BD^2=CD^2$ and $AE^2=CE^2$. Hence $PD^2-PE^2=CD^2-CE^2$, so $P$ also lies on the perpendicular to $DE$ through $C$.

I think that we must say that D, E and F aren't collinear because if they are, the lines through A, B, C perpendicular to EF, FD, DE respectively will be parallel

Let the points where the perpendiculars from $A$,$B$,$C$ to the lines $EF$,$FD$,$ED$ be$K$,$L$,$M$ respectively.......... and the points where $OK$,$OE$,$OD$ meet $AB$,$AC$,$BC$ respectively be $M$,$N$,$P$. (O-circumcentre) using radical axis theorem for the circumcircles of $AKFM$,$BLFM$,$BLDP$,$DPCM$,$DCNE$,$ENAK$ the result follows... Nice and simple!!!!!

Two solutions using PoP and Radical Axis. 1. Just check that the Radical Axes of the cyclic quadrilaterals meet at the Radical Center, our desired point. And it's done! 2. Notice that the power of the desired point wrt the cyclic quads (check 'em out) is same. Thus, it's a definite single point.

This is just Carnot's theorem used. See Kiran Kedlaya's Book Geometry Unbound for precisely the same problem.

My figure might look strange, but try to decode it. The proof becomes too easy. Tell me if I am wrong, because I'm not sure. draw circle to prove AG as radical axis and GH as radical axis. Subsequently we get AH as radical axis. same with others

W cordbash

Let $\omega_A$ be the circle centered at $D$ passing through $B$ and $C$, and define $\omega_B$ and $\omega_C$ similarly. Then, the three lines in question are the pairwise radical axes of $\omega_A,\omega_B,\omega_C$, so we are done.

It is just orthological two triangles and this is trivial....

Another cool problem solved simply by radical center. There isn't much information for us to use angle chasing or "synthetic" techniques like Ceva. Construct circles centered at D passing through B (and consequently C), etc. Then their pairwise radical axes pass through A and perp. to EF, thru B and perp. DF, thru C perp. DE, and by radical center they concur. $\blacksquare$

Let the foot of the perpendicular from $A$ to $EF$ be $X$, the foot of the perpendicular from $B$ to $DF$ be $Y$, and the foot of the perpendicular from $C$ to $DE$ be $Z$. Let $P_1=AX \cap BY$, $P_2=BY \cap CZ$, $P_3=CZ \cap AX$. Since $FXP_1Y, DYP_2Z, EXP_3Z$ are cyclic, their circumcircles concur at the Miquel Point of $\triangle DEF$. Then, we can conclude that $P_1=P_2=P_3$, as desired. $\blacksquare$

^ >>> The desired concurrency point is the radical center of the circle centered at $D$ through $B,C$, at $E$ through $C,A$, and at $F$ through $A,B$. $\square$

slightly different approach

Dear Mathlinkers, a long discussion...about two orthologic triangles here then A propos de deux triangles orthologiques Sincerely Jean-Louis

what radical axis?

Consider the circles $\Gamma_1, \Gamma_2, \Gamma_3$ with respective centres $D, E, F$ and through $B, C, A$ respectively. Then the radical axes of these circles are the required perpendicular lines, and are either parallel or concurrent. (It seems that the case where they're parallel can't be avoided, in fact it is visible when $D, E, F$ are collinear. So I'll presume they meet at the point at infinity for that orientation of parallel lines) $\square$

Construct the circles $\omega_1, \omega_2, \omega_3$ with centers $D, E, F,$ respectively. Then let $X, Y, Z$ be the second point of intersections between $\omega_1 \cap \omega_2 (\neq A), \omega_2 \cap \omega_3 (\neq B),$ and $\omega_3 \cap \omega_1 (\neq C),$ respectively. Now $AY, BZ, CX$ concur at their radical center, which is our desired point of concurrency. $\blacksquare$

Just apply the trigonometric form of Ceva's theorem to $\triangle ABC$ and $\triangle DEF$