This is a China Olympiad Problem, for example see here. [Mod: also here]

Quote: Let $\triangle{ABC}$ be a triangle with orthocenter $H$ . The tangent lines from $A$ to the circle $w$ with diameter $[BC]$ touch it on $P$ and $Q$. Prove that $H\in PQ$ . Proof 1 (pole/polar). Denote the circumcircle $(O)$ of $\triangle ABC$ , $\{A,D_1\}=AH\cap (O)$ , $D\in BC$ for which $AD\perp BC$ and $\{U,V\}=AH\cap w$ . Observe that $DU^2=DB\cdot DC=DA\cdot DD_1=DA\cdot DH$ , i.e. $\boxed {\ DU^2=DA\cdot DH\ }$ what is a characterization of $(A,H,U,V)$ - harmonical division, i.e. $H$ is harmonical conjugate of $A$ w.r.t. $\{U,V\}$ . In conclusion, $H$ belongs to polar of $A$ w.r.t. $(O)$ , i.e. $H\in PQ$ . Proof 2 (metric). Suppose that $AB$ separates $P$ , $C$ . Denote $E\in AC$ so that $BE\perp AC$ and $F\in AB$ so that $CF\perp AB$ . Observe that $\frac {PF}{PB}=\sqrt {\frac {AF}{AB}}$ and $\frac {QE}{QC}=\sqrt {\frac {AE}{AC}}$ . Thus $\frac {PF}{PB}\cdot\frac {QE}{QC}=|\cos A|=\frac {EF}{BC}$ , i.e. $\frac {PF}{PB}\cdot\frac {QE}{QC}=\frac {EF}{BC}$ $\Longleftrightarrow$ $BP\cdot FE\cdot QC=PF\cdot EQ\cdot CB$ $\stackrel{(*)}{\Longleftrightarrow}$ $BE$ , $PQ$ , $FC$ are concurrently $\Longleftrightarrow$ $H\in PQ$ . Lemma. Let $ABCDEF$ be a cyclic hexagon. Prove that $\boxed {\ AD\cap BE\cap CF\ne\emptyset \Longleftrightarrow AB\cdot CD\cdot EF=BC\cdot DE\cdot FA\ }\ (*)$ .

I did it with poles and polars: We are going to use the circle with diameter $BC$ as the circle of reciprocation. Proving that $P,Q,H$ are collinear its the same as proving that their polars are concurrent. Because both $P$ and $Q$ belong to the circle, their polars are the tangents to the circle that meet at $A$. So, we must prove that the polar of $H$ passes through $A$. Let $AE$ be the perpendicular to $HM$($M$ is the midpoint of $BC$). If $E$ is the inverse of $H$, then $AE$ is the polar of $H$ and we are done. So we must prove that $(ME)(HM)={BM}^{2}$. Because $\triangle{AEH} $~$ \triangle MDH$ we have : $\frac{EH}{HD}=\frac{AH}{HM}$ $\Longrightarrow$ $(EH)(HM)=(AH)(HD)$. On the other hand $(ME)(HM)={BM}^{2}$ $\Leftrightarrow$ $(EH)(HM) + {HM}^{2}={BM}^{2}$ $\Leftrightarrow$ $(AH)(HD) + {HM}^{2}={{R}^{2}}{{\sin A}^{2}}$ $\Leftrightarrow$ $4{R^{2}}\cos A\cos B\cos C + HD^{2} + DM^{2}={{R}^{2}}{{\sin A}^{2}}$ $\Leftrightarrow$ $4{R^{2}}\cos A\cos B\cos C$ $+ 4{R^{2}}{\cos B^{2}}{\cos C^{2}} +$ ${{R}^{2}}{{\sin A}^{2}}$ $+ 4{R^{2}}{\cos B^{2}}{\sin C^{2}} - $$4{{R}^{2}}\sin A\cos B\sin C$ $={{R}^{2}}{{\sin A}^{2}} $ Which is easy to verify that is true. QED (Sorry for posting the pole/polar solution again! i was finishing it when it appeared Virgil Niculas' solution)

Let $AD,BE,CF$ the altitudes of $\triangle {ABC}$, and $O$ the midpoint of $BC$. Because $\angle APO=\angle AQO=\angle ADO=90$, the points $A,P,Q,O,D$ are concyclic. Let $\omega$ the circumcircle of $\triangle {APQ}$. Apply an inversion with center at $A$ and radius $|AP|$. Because $AP^2=AH\cdot AD$, we obtain that $H$ is the image of $D$ under this inversion, and $P,Q$ remains invariant. For the other hand, $\omega$ is transformed into the line containing $P,Q$. Because $D\in \omega$, we can conclude that $H\in \overline {PQ}$.

Points $P$ and $Q$ are the intersections of the circles $\mathcal {C}(M,MB)$ and $\mathcal{C}(N, NA)$, $M$ and $N$ being the midpoints of $ BC$ and $AM$ respectively. If $D, E, F$ are the feet of the altitudes on $BC, CA$ and $AB$ respectively and these altitudes concur at $H$, then we know that $AH\cdot DH=BH\cdot HE$, i.e. $H$ has equal power w.r.t. to both circles, consequently it belongs to their radical axis $PQ$. Best regards, sunken rock

Many nice solutions! that's great! thanks to Virgil Nicula for the interesting lemma that I'll try to prove later.

Virgil Nicula's lemma is similar to this: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=360186 except that the points $B,E$ are switched.

oneplusone wrote: Virgil Nicula's lemma is similar to this: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=360186 except that the points $B,E$ are switched. you're right! they are similar, to prove that part ${ \ AD\cap BE\cap CF\ne\emptyset\Longleftarrow AB\cdot CD\cdot EF=BC\cdot DE\cdot FA\ }\ (*) $ from Virgil's lemma, we can also assume that it's wrong i.e there is three intersection point, if I remembre well it's we get that $a.b.c=x.y.z$ with $a>x$ , $b>y$ and $c>z$ which obviously wrong, and then deduct the desired result.

Let $AH \cap PQ = H'$ and $AH \cap (circle) = X, Y$(X is on the same side with A from BC) $(circle) \cap AB = R$ we easily know that $C, H, R$ is collinear and perpendicular to $ AB$ since $AR \perp HR$ and $\angle XRH=\angle XBC=\angle YBC = \angle HRY$ $AXHY$ is harmonic also, it's well known that $AXH'Y$ is harmonic therefore $H=H'$ $P, H, Q$ is collinear since $P, H', Q$ is collinear

Hi everyone .Let me inform you that this problem is from China 1997 and i have three solutions but two of them have already posted so i'll post the third one: According to the following picture its enough to prove that $\hat{AQK}=\hat{AQH}$ Let $AD,BE,CF$ the altitudes of $\triangle {ABC}$, and $O$ the midpoint of $BC$.Since $\hat{APO}=\hat{AQO}=\hat{ADO}=90^{o}$ then the points $A,P,Q,O,D$ are concyclic. We have: $AQ^{2}=AH\cdot AD=AE\cdot AC $ So $\hat{HDQ}=\hat{HQA}=\hat{ADQ}(1)$ and $\hat{APQ}=\hat{ADQ}=\hat{AQH}(2)$ As a result $\hat{APQ}=\hat{AQP}\stackrel{(2)}=\hat{AQH}$ and we are done. [geogebra]9efe53ffcf7728717ef09b8443207d80b9aa5403[/geogebra] PS:The following problem was also posted in China 2005 i think(its almost the same): Let $\triangle{ABC}$ be a triangle and its altitude $AD$.The tangent lines from $A$ to the circle with diametre $BC$ touch this circle on $P$ and $Q$.If $H\equiv AD \cap PQ$ then prove that $H$ is the orthocenter of $ABC$ Best regards, Chris

Obviously points $A,P,M_{a},D,Q$ are concyclic where $M_{a}$ is the midpoint of $BC$. Let's call $\Gamma_{a}$ and $\Gamma_{b}$ circles with diamters $BC$ and $AC$ respectivly. Since $(APQ)\cap\Gamma{a}\equiv(QP)$ and $(APQ)\cap\Gamma{b}\equiv(AD)$ and $\Gamma{a}\cap\Gamma{b}\equiv(CF)$ then by radical axis theorem it follows immediatly that $(QP)\cap(AD)\cap(CF)\equiv H$ and hence the points $P,Q$ and $H$ are collinear.

I have a very short proof.Let $M$ be the midpoint of $BC$.Extend $MH$ to meet $\odot ABC$ at $M'$ where $M'$ and $H$ are on the same side of $M$.Now $A$ is the pole of $PQ$ wrt the circle $(M,MB)$.Again note that $AM'$ is the polar of $H$.So polar of $H$ passes through $A\rightarrow H $lies on $PQ$.Hence proved.

Let the altitudes be $AD$, $BE$, $CF$. Let the foot of the perpendicular from $H$ to $AM$ be $X$. By http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2173258&sid=fd7168f527cac736647f7961645922f0#p2173258, $MX \cdot MA = BM^2$ so the polar of $A$ with respect to the semicircle with diameter $BC$ passes through $X$, and therefore $H$. So $P$, $Q$, $H$ are collinear.

Another solution: Let the three altitudes be $AK, BG$ and $CF$. Easy to see that $P, Q$ lie on the circle $C_{1}$ with diameter $AD$. Let $GF$ cut $BC$ at $L$. $PQ = a$ is the polar of $A$ with respect to $C_{2}$ But from the configuration $LH$ is also the polar of $A$. It follows that $L, P, H, Q$ are collinear.

you can instakill this with poles or polars, but here's a quick radical axis argument. Let $M$ be the midpoint of $BC$, $A_1$ be the foot of $A$ onto $BC$, $B_1$ be the foot of $B$ onto $AC$. We just want to show that $H$ is on the radical axis of the circles with diameters $BC$ and $AM$ (which is $PQ$). But this is obvious, since $(HB)(HB_1)=(HA)(HA_1)$. Note that $\triangle ABC$ is implied to be acute...

By pole and polar. AD,BE and CF are the altitudes of ABC.Draw AL perpendicular on OH (possibly extended) meeting OH at L. Then we have OH*OL=OH*(OH+HL)=OH^2+OH*HL. Now <ADO=<ALO=90 degrees imply that points A,L,D,O are conclyclic. Hence OH*HL=AH*HD, so, OH*OL=OH^2+AH*HD. Substituting OH^2=2R^2[(cosB)^2+(cosC)^2]-a^2/2 , AH=2RcosA and HD=2RcosBcosC and simlifying, we get OH*OL=a^2/4. Thus L is the invese point of O, which implies that the polar of H passes through A. Hence the polar of A must also pass through H. But polar of A is PQ. Hence the result.

Dear Mathlinkers, 1. by definition, PQ is the polar of A wrt the circle with diameter BC 2. B', C' being the feet of thze B, C-altitudes of ABC are on the last circle 3. H being the point of intersection of BB' and CC' is on PQ (result of de La Hire Conics I prop. 22, 23 I will put this last resul this week... Sincerely Jean-Louis Ayme J.-L., La réciprocité de Philippe de La Hire, G.G.G. vol. 13, p. 21-25 ; http://perso.orange.fr/jl.ayme

Quick with projective. Let the circle with diameter $BC \equiv \Omega$ and $BH \cap AC \equiv E$. Observe that $PQ$ is the polar of $A$ wrt $\Omega$. Now $\angle XEH = \angle XYB=90-\angle CYX=\angle BCY = \angle BEY$, and $\angle BEA=90$. Thus $(A, H; X, Y)=-1$ so the result follows.

$A$ lies on the polar of $H$ wrt the circle on diameter $\overline{BC}$ (Brokard) so $H,P,Q$ are collinear by La Hire.

Let $X, Y$ be the intersections of the $A$-altitude with the circle. $H, P, Q\text{ collinear}\Leftrightarrow H\in \text{polar}(A)\Leftrightarrow (A, H; X, Y)=-1\Leftrightarrow(X, Y; B,C)=-1\Leftrightarrow BX\cdot CY = BY\cdot CX$ which is obvious.

Let $E,F$ be the feet of perpendiculars from respectively $C,B$ take an inversion with $A$ as the center and radius $AP$. This inversion takes $P,Q$ to themselves and it takes $H$ to feet of perpendicular from $A$ to $BC$ ($\angle AEH=\angle AH^{'}E^{'}=\angle AH^{'}B=\frac{\pi}{2}$)thus the images of $H,P,Q$ and $A$ are concyclic hence the conclusion

I solve this problem with pole and polar.First we can note $E$ is on the cicrcle such as the $BE$ is perpendicular to $AC$ and analogus the point F. Now we have 4 points on the circle $B,C,E,F$ and $BF\cap CE={A}$ and we have from theorem of Hire that the intersection point of BE and CF is on the polar of the point A which is PQ so P,Q and H are collinear.

Does anybody have a solution with complex numbers?

very short

@Analgin, Set $b=-1,c=1$. Then let $p,q$ be points on the unit circle, so $a=\frac{2pq}{p+q}$. Then if $DEF$ is the orthic triangle, $$e=\frac{p+q-2pq}{2-(p+q)},f=\frac{p+q+2pq}{2+p+q}.$$Now with the chord intersection formula, $$h=\frac{e+2ef-f}{e+f}=\frac{2p^2q^2-p^2-q^2}{(p+q)(pq-1)}.$$It remains to compute $$\frac{p-h}{p-q}=\frac{p-\frac{2p^2q^2-p^2-q^2}{(p+q)(pq-1)}}{p-q}=\frac{q(p^2-1)}{(p+q)(pq-1)}=\frac{q-pq\overline{h}}{q-p}=\overline{\left(\frac{p-h}{p-q}\right)},$$so $P,H,Q$ are collinear, as required.

An easy extension. Let $\alpha=\mathbb C(O)$ be circumcircle of $\triangle ABC$ and $w=\mathbb C(K)$ be a circle so that $\{B,C\}\subset w$ and exists $D\in (BC)$ such that $DA\perp DK\ .$ Let $\{A,D_1\}=\{A,D\}\cap \alpha$ and the symmetric $L$ of $D_1$ w.r.t. $D\ .$ The tangent lines from $A$ to the circle $w$ touch it on $P$ and $Q\ .$ Prove that $L\in PQ$ (see PP10 from here).

jlammy wrote: @Analgin, Set $b=-1,c=1$. Then let $p,q$ be points on the unit circle, so $a=\frac{2pq}{p+q}$. Then if $DEF$ is the orthic triangle, $$e=\frac{p+q-2pq}{2-(p+q)},f=\frac{p+q+2pq}{2+p+q}.$$Now with the chord intersection formula, $$h=\frac{e+2ef-f}{e+f}=\frac{2p^2q^2-p^2-q^2}{(p+q)(pq-1)}.$$It remains to compute $$\frac{p-h}{p-q}=\frac{p-\frac{2p^2q^2-p^2-q^2}{(p+q)(pq-1)}}{p-q}=\frac{q(p^2-1)}{(p+q)(pq-1)}=\frac{q-pq\overline{h}}{q-p}=\overline{\left(\frac{p-h}{p-q}\right)},$$so $P,H,Q$ are collinear, as required. Sorry i couldn't understand your solution can you explain it more detailed especially first part

As usual, the coordinate bash is not hard.

Quote: Let $\triangle{ABC}$ be a triangle with orthocenter $H$ . The tangent lines from $A$ to the circle $w$ with diameter $[BC]$ touch it on $P$ and $Q$. Prove that $H\in PQ$ . I can't understand the reason before such long long long long solutions. Let \(M\) denote the mid of segment \(BC.\) Clearly, \(A, P , H , M , Q\) are concyclic in that order. Also, \(Pow_\omega H = Pow_{APQ} H\)..Therefore \(H\) must lie on the radical axes of the circles \(\omega\) and \(APQ\) which clearly is the claimed line the \(PQ\).

Dear Mathlinkers, You can have a look at http://jl.ayme.pagesperso-orange.fr/Docs/Quickies%205.pdf p.24-25 Sincerely Jean-Louis

@Anar24 Suppose BCPQ is inscribed in unit circle. Because BC is diameter jlammy supposed that $b=-1,c=1$. Then he used formula for tangents intersection. Next he used formula for foot $E$ of $C$ on $AB$

Virgil Nicula wrote: Lemma. Let $ABCDEF$ be a cyclic hexagon. Prove that $\boxed {\ AD\cap BE\cap CF\ne\emptyset \Longleftrightarrow AB\cdot CD\cdot EF=BC\cdot DE\cdot FA\ }\ (*)$ . Is this lemma true for not convex hexagons? If it's not true, then your proof doesn't work in case $H$ lies outside triangle $ABC$ in given problem.

The polars of $P,Q$ wrt $(BC)$ both pass through $A$, so by La Hire it suffices to prove that the polar of $H$ does as well. Let the reflection of $H$ over the midpoint $M$ of $BC$ be $H_1$, and $HM\cap (ABC)=Q$; we can see that $$MH\cdot MQ=MH_1\cdot HQ=MB\cdot MC$$by orthocenter reflection, and since $Q$ is the $A$-queue point, we know that $AQ\perp QH$, which implies that $AQ$ is the polar of $H$.

JBMO2000 Problem 3 https://artofproblemsolving.com/community/u229790h6143p27113536

Consider the circle with center $A$ and radius $AP$. Then the problem is equivalent to showing that $H$ lies on the radical axis of this circle and the circle with diameter $BC$. We can do this by showing $H$ has equal powers with respect to both circles. This resolves to showing $AH^2 - AP^2 = MH^2 - MP^2$ or $AH^2 - HM^2 = AP^2- MP^2$. Notice that we have $AP^2 = AM^2 - MP^2$, and $AM^2 = \frac{AB^2 + AC^2}{2} - MB^2$, giving $AP^2-MP^2 = AM^2 - 2MB^2 = \frac{AB^2 + AC^2}{2} - 3MB^2$. Then let the foot from $A$ to $BC$ be $D$. We can then write $AH^2 - HM^2 = AH^2 - (CH^2 - CD^2 + CM^2) = AH^2 - (CH^2 - CD^2 + (CD - MC)^2) = AH^2 - (CH^2 - 2CD \cdot MC + MC^2) = AH^2 - CH^2 + CD \cdot BC + MC^2$. We then want to show $AH^2 - CH^2 + CD \cdot BC - MB^2 = \frac{AB^2 + AC^2}{2} - 3MB^2$, which we can reduce to $AH^2 - CH^2 + CD \cdot BC = \frac{AB^2 + AC^2 - BC^2}{2} $. Then by Law of Cosines, we have $AH^2 = CH^2 + AC^2 - 2CH \cdot AC \cdot \cos{ACH}= CH^2 + AC^2 - 2CH \cdot CE$ where $E$ is the foot from $C$ to $AB$. We can then write $AH^2 - CH^2 = AC^2 - 2CH \cdot CE$. By Power of a Point on $(BEDH)$, we have $CH \cdot CE = CD \cdot BC$. We then just need to show $AC^2 - BC \cdot CD = \frac{AB^2 + AC^2 - BC^2}{2} $. Multiply everything by two to get the equivalent $2AC^2 - 2BC \cdot CD = AB^2 + AC^2 - BC^2$ and rearranging gives the equivalent $AC^2 + BC^2 - 2BC \cdot CD = AB^2$, but $CD = AC \cdot \cos C$, so we are done by LOC.

Let $D$ be the altitude from $A$ to $\overline{BC}$ and $O$ the circumcenter of $(ABC)$. Also denote by $\measuredangle$ the angle $\angle$ modulo $\pi$. Then we make the following claim: Claim: $A, P, D, O, Q$ is cyclic. Proof: Indeed we have that \[\measuredangle APO = \measuredangle AQO = \measuredangle ADO = 90^{\circ}. \blacksquare\] Now observe that if we consider circles $(BFHD)$ and $(DHQC)$, that $A$ is the radical center of all of our present circles so that \[AP^2 = AQ^2 = AH \cdot AD \implies \triangle AHP \sim \triangle APD; \phantom{c} \triangle AHQ \sim \triangle AQD.\] Now the solution is immediate: \begin{align*} \measuredangle AHP &= \measuredangle DPA \\ &= \measuredangle DQA \\ &= \measuredangle AHQ, \end{align*}so that $P, H, Q$ are collinear, as desired. $\blacksquare$

short, nice

Let $M$ be midpoint of $BC$, then just check $X=AM \cap PQ$ is Humpty, since $MB^2=MP^2=MX*MA=MC^2$, from which the result follows.

Let $\Delta DEF$ be the orthic triangle, $M$ be the midpoint of $BC$. Note that $AP^2=AQ^2=AB \cdot AE = AH \cdot AD$. Also, $D,P,Q$ clearly lie on the circle with diameter $AM$. Take an inversion centred at $A$ with radius $\sqrt{AH \cdot AD}$, then $P$ and $Q$ go to themselves, and $H$ goes to $D$. Since $A,D,P,Q$ are concyclic, inverting back gives that $H,P,Q$ collinear as desired. $\square$