Beautiful Problem!

Well-known problem embedded WRT $\Delta AH_BH_C$ India Practice TST 2019 #2 P1 wrote: Let the points $O$ and $H$ be the circumcenter and orthocenter of an acute angled triangle $ABC.$ Let $D$ be the midpoint of $BC.$ Let $E$ be the point on the angle bisector of $\angle BAC$ such that $AE\perp HE.$ Let $F$ be the point such that $AEHF$ is a rectangle. Prove that $D,E,F$ are collinear. Solution: Let $H_B,H_C$ be foot from $B,C$. Note, $\odot (AEHF)$ $\equiv$ $\odot (AH_BH_C)$ $\implies$ $AF$ is exterior bisector WRT $\angle BAC$, $E$ is midpoint of arc $H_BHH_C$ and $D$ is intersection of tangents at $H_B,H_C$ $\implies$ $D,E,F$ collinear

Do an inversion around $A$ with radius $\sqrt{AH.AX}$ where $X$ is the foot of perpendicular of the $A$ on $BC$. Then $$E \mapsto AE \cap BC, F \mapsto AF \cap BC$$and $D \mapsto (AD) \cap \odot(BHC)$ which is the $A-$HM point in $\triangle ABC$. But they lie on the A-apollonious circle. So done.

Why so complicated? This is from Bosnia and Herzegovina: https://artofproblemsolving.com/community/c6h1709471p11016808

Too Easy for a TST Problem but nice... Indian TST Practise Test 2 P1 wrote: Let the points $O$ and $H$ be the circumcenter and orthocenter of an acute angled triangle $ABC.$ Let $D$ be the midpoint of $BC.$ Let $E$ be the point on the angle bisector of $\angle BAC$ such that $AE\perp HE.$ Let $F$ be the point such that $AEHF$ is a rectangle. Prove that $D,E,F$ are collinear. Solution:- Let $M$ be the intersection of the diagonals $AH$ and $FE$. Let $X,Y,Z$ be the feet of altitudes from $A,B,C$ respectively and let $FE\cap ZY=K$. Notice that $FE\perp ZY$ as $\angle KEZ+\angle KZE=\angle FAZ+\angle ZAE=90^\circ\implies FE\perp ZY$. Now by Radical Axis on $\odot(AZHY)$ and $\odot(BZYC)$ we get that $MD\perp ZY$. This forces $D-E-F$. $\blacksquare$.

Easy and nice. Indian TST 2019 Practice Test 2 P1 wrote: Let the points $O$ and $H$ be the circumcenter and orthocenter of an acute angled triangle $ABC.$ Let $D$ be the midpoint of $BC.$ Let $E$ be the point on the angle bisector of $\angle BAC$ such that $AE\perp HE.$ Let $F$ be the point such that $AEHF$ is a rectangle. Prove that $D,E,F$ are collinear. Let the feet of the altitudes from $A,B,C$ to sides $BC,CA,AB$ be $P,Q,R$ respectively. $\angle AFH = \angle AQH = 90 \Longrightarrow A,F,P,E,Q$ are concyclic with diameter $EF = AH$. By Fact 5, $EP = EQ \Longrightarrow FP = FQ$ which implies that $EP \perp PQ$. Thus, $EE\parallel PQ \parallel FF \Longrightarrow FPEQ$ is harmonic. Hence, $E,F,D$ are collinear.

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ayan.nmath wrote: Let the points $O$ and $H$ be the circumcenter and orthocenter of an acute angled triangle $ABC.$ Let $D$ be the midpoint of $BC.$ Let $E$ be the point on the angle bisector of $\angle BAC$ such that $AE\perp HE.$ Let $F$ be the point such that $AEHF$ is a rectangle. Prove that $D,E,F$ are collinear. As the title suggests its really easy.. Anyways i will post my solution then also. Let $H_A, H_B, H_C$ be the Base of altitude from $A, B, C$ Respectively. Now Just observe $AH_BH_CE$ is cyclic With Circumcenter $X$ Then as $\angle HEA=90^\circ$ So $X\in AH$ And also as $E\in$ Angle bisector of $\angle BAC$ So $H_BE=H_CE$ Now just observe if $D$ is the midpoint of $BC$ then $H_BD=H_CD=\frac{BC}{2}$ and also $\angle XH_BD=\angle XH_CD=90^\circ$ Hence $H_BD, H_CD$ are tangents from $D$ and as $H_BE=H_CE$ So clearly $X, E, D$ Are collinear. And as $F\in XE$ So $F, E, D$ Are collinear $\blacksquare$ Also Here's More harder Version of the question.. Why in the question $O$ is Defined??

This is India TST Problem? Shame. It is great configuration though. Let X and Y be the foot of perpendicular from B to AC and C to AB respectively. We see that due to Three Tangents Lemma, DX and DY are tangent to circle with diameter AH and so D lies on perpendicular bisector of segment XY and E and F also lie on this circle due to the given conditions, now we see that AE is internal angle bisector of \angle XAY so E lies on perpendicular bisector of segment XY since E lies on circle with diameter AH and since \angle FAE = 90^\circ, we see that F lies on external angle bisector of \angle XAY and so F lies on perpendicular bisector of segment XY as F lies on circle with diameter AH, so points D, E, F lie on the perpendicular bisector of segment XY as desired PS : By today I will be allowed to LaTeX this post so for now posting a non-LaTeXed solution

This can be generalized and its not only for the orthocenter $\textrm{Generalized Form}$ : Let $P$ be the isogonal conjugate of any point on the perpendicular bisector of $BC$ , let $E , F$ be the feets of the perpendicular from $P$ to the internal and external bisectors of angle $\angle{BAC}$ , Prove that $E , F$ and the midpoint of side $BC$ are collinear!

Wait why are the previous solns so complex? Let $N$ be the midpoint of $AH$ and let $X,Y$ be the altitudes to $AC,AB$ $N$ is the center of $(AXHY)$ and also its well known that $DX,DY$ are tangent to this circle. So, $DN$ intersects the circle at the midpoints of the minor and major arcs $XY$. But since in $\triangle AXY$, $AE$ is angle bisector, it passes through the midpoint of the minor arc and we easily see that $F$ is the midpoint of major arc and so we're done

RamtinVaziri wrote: This can be generalized and its not only for the orthocenter $\textrm{Generalized Form}$ : Let $P$ be the isogonal conjugate of any point on the perpendicular bisector of $BC$ , let $E , F$ be the feets of the perpendicular from $P$ to the internal and external bisectors of angle $\angle{BAC}$ , Prove that $E , F$ and the midpoint of side $BC$ are collinear! Can someone please tell how to prove this generalization.

Since $AEHF$ is a rectangle, we know $EF$ meets $AH$ at its midpoint $M$. It's well-known that $DM \parallel AO$. Because $E, M, F$ are collinear, it suffices to show $E \in DM$. Clearly, $M$ is the circumcenter of $(AEHF)$, and $AH$ and $AO$ are isogonal wrt $\angle BAC$. Hence, we have $$\angle OAE = \angle HAE = \angle MAE = \angle MEA$$implying $EM \parallel AO$, which suffices. $\blacksquare$

Wew Let $M$ be the midpoint of $AH$ and $H_B,H_C$ be the feet of perpendicular from $B,C$ onto the opposite sides respectively. $A,F,H_C,H,E,H_B$ lies on the circle with diameter $AH$. Note that $E$ is the midpoint of minor arc $H_BH_C$ so $F,M,E$ lies on the perpendicular bisector of $H_BH_C$. $DM$ is also perpendicular to $H_BH_C$ which is the radical axis of $\{(AH_BH_C) , (BH_CH_BC)\}$. Combining all of them we get $F-M-E-D$ are collinear.

Invert at $A$ sending $H$ to $AH\cap BC$. Since the $A$-Humpty point lies on the $A$--apolonius circle we're done.

Let $I$ be the midpoint of AH. Diagram attached for reference Introduce the nine point circle. Notice that $\overline{F-I-E}$ by a basic property of rectangles. So this is equivalent to proving $\overline{I-E-D}$. Unfortunately this dies to simple angle chasing, if I’m not high. We shall prove that $\angle{EIH}=\angle{DIH}$.$\angle{EIH}=2\angle{IAE}$ because $I$ is the midpoint of the hypotenuse. Therefore $\angle{EIH}=A+2B-180=B-C$. Now to calculate $\angle{DIG}=\angle{DLG}=\angle{DLB}-\angle{GLB}=B-C$ since $D$ is the midpoint of he hypotenuse of another right angled triangle.

Cute but easy problem Let $AH \cap FE=K$ and as diagonals bisect each other $\Rightarrow K$ is the mid-point of $AH$ Let $X,Y$ be the foot of altitudes on $AC$ and $AB$ respectively. Then $X,Y \in \odot(AEHF)$ with $K$ as the circumcentre. Constructing a nine point circle of $\triangle ABC$,we see that $XY$ is the radical axis of $\odot(AEHF)$ and $\odot(DXY)$ It's also well know that $DX$ and $DY$ are tangents of $\odot(AEHF)$ at $X$ and $Y$. This is enough to imply $D$ lies on the perpendicular bisector of $XY$ and so does $E$ because $AE$ is the angle bisector of $\triangle AXY$. Also $K$ lies on the perpendicular bisector and hence $K-E-D$ is collinear implying $F-E-D$ collinear.

Let the foot of altitude from $B$ and $C$ be $H_B$ AND $H_C$. By Three Tangent lemma, $DH_B$ and $DH_C$ are tangent to $(AH_CHH_B)$. So $DH_B=DH_C$ Also $AE$ bisects $\angle BAC$. Hence by Incenter Excenter lemma, $E$ is the Mid-point of the minor arc $H_BH_C$. So $H_BE=H_CE$. As $\angle FH_BE=\angle FH_CE=90^o$, by congruent triangles, $FH_B=FH_C$. Hence $(FH_CEH_B)$ is harmonic and the result immediately follows. $\blacksquare$

gambi wrote:

beautiful

It's almost $1$ linear with projective, why anyone didn't try to this? Let $\omega$ and $\Gamma$ be be circumcircles of $\triangle ABC$ and $\triangle AHE$, respectively. Let $AE\cap \omega=T, TD\cap \omega= T', BH\cap AC=B', CH\cap AB=C'$. Since $\angle FAE=\angle T'AE=90$, $F-A-T'$ are collinear. Since $BTCT'$ is kite, $(B,C;T,T')=-1$ and projecting this to $\Gamma$ with pencil $A$, we get $(F,E;B',C')=-1$. Since $DB'$ and $DC'$ are tangents to $\omega$, we get $E-F-D$ are collinear.

Dear Mathlinkers, this nice problem comes from Droz-Farny... http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=481340 Sincerely Jean-Louis

Let $M$ be the mid-point of $\overline{AH}$. Since $\angle AE\perp HE$, $\overline{AH}$ is the diameter of $(AEHF)$. So, $M$ is the center of $(AEHF)$. So, $E,M,F$ are collinear. Let $B_1$ and $C_1$ be the feet of perpendiculars of $B$ and $C$ onto their opposite sides, respectively. Then, we $A,B_1,H,C_1$ are concyclic, so $B_1,C_1\in(AEHF)$. Let $t$ be the perpendicular bisector of $\overline{B_1C_1}$. Then, $M,D\in t$. Moreover, $\angle EAC=\angle BAE=\angle EAB_1{\stackrel{B_1\in(AEC_1)}{=}}\angle EC_1B_1=\angle C_1AE{\stackrel{C_1\in(AEB_1)}{=}}\angle C_1B_1E$. So, $\angle EC_1B_1=\angle C_1B_1E$, giving $\triangle B_1EC_1$ is isosceles triangle. So, $E\in t$. But since $M\in t$ and $M\in\overline{EF}$, it means $F,M,E,D\in t$, as desired. $\square$

$\text{It's not that complicated :)}$

Here's a soln using Nine-Point Circle Configuration!! Let the midpoint of $AH = X$ , the midpoint of $OH = N =$ nine-point center Then we know $DX$ is a diameter of Nine point circle. So $D-N-X$ are collinear $AEHF$ is rectangle $\implies$ $EF$ bisect $AH\implies X \in EF$ By isogonal conjugates, and angle chasing: $AO \parallel EF$ By midpoint theorem: $EF$ bisects $OH \implies N \in EF$ $\implies D \in EF$

Let $XYZ$ be the orthic triangle of $\Delta ABC$ Clearly $E$ and $F$ are on $(AYHZ)$ I know because of EGMO that $DZ$ and $DY$ are tangents to aforementioned circle. (proving is pretty simple if you know your configs, $\astrosun (BZYC) = D$ so you get $\angle YDZ=2\angle YBD= \pi - 2A$ and because $DZ=DY$ we get $\angle DZY = A = \angle ZAY$) Then $\angle ZAE = \angle YAE$ so $E$ is arc midpoint of $YZ$ and as $F$ is antipode of $E$, it is also an arc midpoint of $YZ$, hence, $EF$ and the tangents ay $Y$ and $Z$ concur... really fun problem to do at 4:30 in the morning, life is good : )

Let $AH \cap \odot(ABC)=S,AE\cap \odot(ABC)=T,R$ is the reflection of $T$ by $BC$ Notice $\angle HRT=\angle STR=180^{\circ}-\angle HST=\angle AHR$ Which means that $H-E-R$ So $\angle TER=90^{\circ},RD=DT$ Hence $DE=DT$ Consider that $RT//AH,FH//ET$ So $\angle FHA=\angle ETD$ Since $\angle FHA=\angle AEF$,and $ED=DT$ Hence $\angle AEF=\angle TED$ Which means that $D-E-F$

Since NBWC is harmonic, we project it through A onto (AH) and get that EF and the bases of heights form a harmonic. Hence EF bisects BC.

where is $O$ used huh Let $M$ be the midpoint of $\overline{AH}$. It clearly suffices to show that $M,E,D$ collinear. Let $P,Q$ be the feet of the altitudes from $B$ and $C$ respectively, so $AEHPQ$ is cyclic with center $M$, so $PM=QM$. Since $E$ is the midpoint of minor arc $PQ$ in $(AEHPQ)$, we have $PE=QE$. Since $BCPQ$ is cyclic with center $D$, we have $PD=QD$. Hence $M,E,D$ lie on the perpendicular bisector of $\overline{PQ}$. $\blacksquare$

Let $X$ be the midpoint of $AH$ , then sps $\angle{HXE}=\theta$ since $FE$ and $AH$ are length of diagonals and $H$ is there midpoint we have $\angle{XEH}=90^{\circ}-\frac{\theta}{2}$ , so $\angle{XEA}=\frac{\theta}{2}$ as $AFHE$ is a rectangle , so $\angle{XAE}=\frac{\theta}{2}$ as $XA=AE$, also $\frac{\theta}{2}=\frac{B-C}{2}$ which means $\theta=B-C$. $\textcolor{blue}{\mathrm{Claim:}} \angle{DXH}=\angle{EXH}$ $\textcolor{blue}{\mathrm{Pf:}}$ clearly $X$ lies on nine point circle of $\triangle{ABC}$ , so it is well known that $XD|| AO$ , so $\angle{HXD}=\angle{HAO}=B-C \implies \angle{DXH}=\angle{EXH}$ $\square$. which means $\overline{X-E-D}$ but also since $AFHE$ is a rectangle so $\overline{F-X-E} \implies \overline{F-E-D}$. $\blacksquare$

Take the circumference of diameter $AH$. Now by taking $H_B$, $H_C$ as the feet of the altitudes through $B$ and $C$ respectively, we get that $E$ and $F$ are in its perpendicular bisector. As it is well-known that $D$ also is in it (which follows from the fact that $DH_B$ and $DH_C$ are tangents to the said circumference), we're done.

This is actually quite a nice complex bash. We set $(AH)$ be the unit circle. Let $P_B,P_C$ be the feet of the altitudes from $B$ and $C$. Then, we set $a=x^2,p_b=y^2$ and $p_c=z^2$. It follows that $h=-x^2$. Then, $E$ is clearly the minor arc midpoint of $P_BP_C$ which means $e=-yz$. Thus, $f=yz$ ($F$ is the reflection of $E$ over the center as we construct $F$ such that $AFHE$ is a rectangle). Then, $D$ is well known to be the intersection of the tangents to $(AH)$ at $P_B$ and $P_C$. Thus, $d=\frac{2y^2z^2}{y^2+z^2}$. Then, \begin{align*} \frac{d-e}{f-e} &= \frac{2y^2z^2}{y^2+z^2}\\ &= \frac{yz}{y^2+z^2} \end{align*}Also, \[\overline{\left(\frac{d-e}{f-e}\right)} = \overline{\left(\frac{yz}{y^2+z^2}\right)}\]Now, note that since $yz$ (which is known to be the complex number denoting the arc midpoint of $P_BP_C$ including $A$), $y^2$ and $z^2$ all lie on the unit circle, \begin{align*} \overline{\left(\frac{yz}{y^2+z^2}\right)} &= \frac{\frac{1}{yz}}{\frac{1}{y^2}+\frac{1}{z^2}}\\ &= \frac{1}{yz} \cdot \frac{y^2+z^2}{y^2z^2}\\ &= \frac{yz}{y^2+z^2} \end{align*}Thus, $\frac{yz}{y^2+z^2} = \overline{\left(\frac{yz}{y^2+z^2}\right)}$ which implies that, $\frac{yz}{y^2+z^2} \in \mathbb{R}$ and thus, $D-E-F$ as needed.

RamtinVaziri wrote: This can be generalized and its not only for the orthocenter $\textrm{Generalized Form}$ : Let $P$ be the isogonal conjugate of any point on the perpendicular bisector of $BC$ , let $E , F$ be the feets of the perpendicular from $P$ to the internal and external bisectors of angle $\angle{BAC}$ , Prove that $E , F$ and the midpoint of side $BC$ are collinear! Nice Let $D$ be point of perpendicular bisector of $BC$ and $P$ its isogonal conjugate. Let $\triangle XYZ$ be Pedal triangle of $P$ with $X,Y,Z$ lie on $AC,AB,BC$ respectively. $$\measuredangle {CBD}=\measuredangle{PBA}=\measuredangle {PZY}= \measuredangle {DCB}=\measuredangle{ACP}=\measuredangle {XZP}$$give us $ZP$ is angle bisector of $\angle XZY$. As $D$ and $P$ are isogonal conjugate, By well know property circumcircle of pedal triangle of both point is same hence $M$ lie on $(XYZ)$. Where $M$ is midpoint of $BC$. Observe that $MX=MY$ (by considering $ZP$ as angle bisector). Hence $M$ lie on perpendicular bisector of $XY$. Now as $AE$ and $AF$ are angle bisector we have $EF$ is perpendicular bisector of $XY$.hence $E,F,M$.

Let $S$ and $T$ be the feet of perpendiculars from $B$ and $C$ to the sides $AC$ and $AB$ respectively. $(AFTHES)$ is cyclic with diameter $AH$ and $AH$ and $EF$ intersect at the midpoint of $AH$, i.e., the center of the circle, since $AHEF$ is rectangle. Since, $EF$ is a line passing through the center of the cirlce $(ASHT)$ and it is also parallel to $AO$, which is in turn perpendicular to $ST$, line $EF$ is actually the perpendicular bisector of $ST$. It is also well known, that $DS$ and $DT$ are tangents to $(ASHT)$, implying $D$ also lies on the perpendicular bisector of $ST$, yielding the collinearity.

Let $B_1$ and $C_1$ be the feet of the $B$ and $C$ altitudes in $\triangle ABC$. Since $\overline{EF}$ is a diameter of $(AH)$ and $EB_1 = EC_1$, we have $(B_1, C_1; E, F) = -1$. Since the tangents to $(AH)$ at $B_1$ and $C_1$ intersect at $D$ by three tangents lemma, $EF$ must also pass through $D$.