geometry
incircle
homothety
circumcircle
Triangle
IMO
imo 2000

Let the reflection of $H_2 H_3$ in $T_2 T_3$ be $l_1$ etc.
Let $M$, $N$ be the intersections of $BI$ and $CI$ respectively with $T_{2}T_{3}$.  Let $M^\prime$ be the point on $H_{2}H_{3}$ such that $M^\prime \hat{M} H_2 = T_2 \hat{M} H_2$.  Let $T_{2}^{\prime}$ be the reflection of $T_2$ in $BI$.  Clearly, this lies on the incircle.
$T_2 \hat{M} I = A\hat{T} _3 T_2 - A\hat{B}M = 90-\frac{A}{2}-\frac{B}{2}=\frac{C}{2}=T_2 \hat{C} I$
$\therefore IT_2 MC$ is cyclic.  Since $I \hat{T_2}C=90$, $BMC = 90$.  So $BH_3 N H_2 M C$ are concyclic.
$\therefore M^\prime \hat{H_2} M = H_3 \hat{B} M = \frac{B}{2} = M \hat{B} C = M \hat{H_2} T_2$
$\therefore MM^\prime H_2 \equiv alent MT_2 H_2$
$\therefore MM^\prime = MT_2$

Also, $T_2 ^\prime \hat{M} T_2 = 2B\hat{M}N=C=2N\hat{M}H_2 = T_2 \hat{M} M^\prime$

This means that $T_2 ^\prime$ is the reflection of $M^\prime$ in $T_2 T_3$, so that $T_2$ is the intersection of $l_1$ with the incircle.  Similarly, $T_2$ is the intersection of $l_3$ with the incircle.  This implies the result.

Here's a different, probably less tricky approach.

Let $A'B'C'$ be the image of $ABC$ under the negative homothety taking the circumcircle to the incircle. It suffices to show that $A'B'C'$ reflects across $T_{1}T_{2}, T_{2}T_{3}, T_{3}T_{1}$ to the orthic triangle.

First, note that $B'C'$ is parallel to $BC$, and $AT_{3}= AT_{2}$, so the reflection of $B'C'$ across $T_{2}T_{3}$ is antiparallel to $BC$. Using analogous facts for the other sides, the three reflections are some homothety of the orthic triangle; we need only show that this homothety is identity.

Now let $l_{B}$ and $l_{C}$ be respective reflections of $A'C'$, $A'B'$ across $T_{1}T_{3}$, $T_{1}T_{2}$. Note that $T_{1}$ is equidistant from $A'B'$ and $A'C'$ since it is the midpoint of the arc $B'C'$. Then it must also be equidistant from $l_{B}$ and $l_{C}$.

Furthermore, since $l_{B}$ is antiparallel to $AC$, the angle $(l_{B}, BC)$ is $A$. Similarly, $(l_{C}, BC) =-A =-(l_{B}, BC)$. Hence $BC$ is parallel to an angle bisector of $l_{B}, l_{C}$, and at the same time $T_{1}$ lies on one of the angle bisectors of $l_{B}, l_{C}$. We quickly verify that $l_{B}$ and $l_{C}$ cut $BC$ on the same side of $T_{1}$, so it follows that $BC$ is an angle bisector of $l_{B}, l_{C}$, and so it contains $l_{B}\cap l_{C}$.

Using the corresponding facts for the other two pairs of lines, we conclude that our triangle is a homothety of the orthic triangle which keeps its vertices on the same sides of $ABC$. It then follows that the homothety is centered around $A$, $B$, and $C$ at the same time, so it must be identity, finishing the proof.

Let $A'B'C'$ be the triangle such that $A'B' \| AB$, $B'C' \| BC$, $C'A' \| CA$ and $A'$, $B'$ and $C'$ lie on the incircle of $ABC$. We claim that $A'B'C'$ is the triangle that results from the described reflections. To show this result we first will prove the following lemma.

[b]Lemma:[/b] In an acute-angled triangle $ABC$ with circumcircle $\Gamma$, $D$, $E$ and $F$ are the midpoints of major arcs $\widehat{BC}$, $\widehat{AC}$ and $\widehat{AB}$, respectively. If $\omega$ is the tangent to $\Gamma$ at $A$, $\ell_C$ is the reflection of $AB$ across $DE$ and $\ell_B$ is the reflection of $AC$ across $DF$, then $\omega$, $\ell_B$ and $\ell_C$ are concurrent.

[i]Proof:[/i] Denote $\angle{A}=a$, $\angle{B}=b$ and $\angle{C}=c$. Let $P$ and $Q$ be the intersections of the perpendicular bisector of $AD$ with $AB$ and $AC$, respectively, and let $\ell_B$ and $\ell_C$ intersect at $R$. Now consider the reflection $P'$ of $P$ across $DE$. Since $E$ is the midpoint of major arc $\widehat{AC}$, $\angle{ADE}=\angle{ACE}=90^\circ - b/2$. Further, since $D$ is the midpoint of major arc $\widehat{BC}$, it follows that $AD$ is the external bisector of $\angle{A}$. This combined with the fact that $PA=PD$ implies that $\angle{PDA}=\angle{PAD}=90^\circ - a/2$. Therefore if $M$ is the intersection of $DE$ with $PP'$, $\angle{PDM}=180^\circ - \angle{PDA}-\angle{ADE}=90^\circ - c/2$. Since $P'$ is the reflection of $P$ across line $EDM$, $DM$ is the bisector of $\angle{PDP'}$ and hence $\angle{PDP'}=2\angle{PDM}=180^\circ - c$. By the same argument, if $Q'$ is the reflection of $Q$ across $DF$, it follows that $\angle{QDQ'}=180^\circ - b$. Since $P$ and $Q$ lie on the perpendicular bisector of $AD$ and $AD$ is the external bisector of $\angle{A}$, it follows that $\angle{PDQ}=180^\circ - a$ and that $PD=QD$. Now since $P'D=PD=QD=Q'D$ and $\angle{PDQ}+\angle{P'DP}+\angle{Q'DQ}=180^\circ - a + 180^\circ - b + 180^\circ - c = 360^\circ$, it follows that $P' = Q'$. Further, since $P$ and $Q$ lie on lines $AB$ and $AC$, respectively, $P'$ and $Q'$ lie on $\ell_C$ and $\ell_B$, respectively. Hence $P' = Q' = R$. As previously established, $\angle{MDR}=90^\circ - c/2=\angle{DFE}$ which implies that $DR$ is tangent to $\Gamma$ by tangent-angle theorem. Hence $\omega$, $\ell_B$ and $\ell_C$ concur at $R$. $\blacksquare$

Now note that since $B'C' \| BC$  and $BC$ is tangent to the circumcircle of $A'B'C'$ at $T_1$, it follows that $T_1$ is the midpoint of major arc $\widehat{B'C'}$. By the same argument $T_2$ and $T_3$ are the midpoints of major arcs $\widehat{A'C'}$ and $\widehat{A'B'}$, respectively. Let $k_A$, $k_B$ and $k_C$ denote the reflections of $B'C'$, $A'C'$ and $B'C'$ across $T_2 T_3$, $T_1 T_3$ and $T_2 T_3$, respectively. By the lemma, the pairwise intersections of $k_A$, $k_B$ and $k_C$ lie on $BC$, $AC$ and $AB$. Let these points be $X$, $Y$ and $Z$, respectively. Since $T_2 T_3$, $T_1 T_3$ and $T_1 T_2$ are perpendicular to the bisectors of $\angle{A}$, $\angle{B}$ and $\angle{C}$, $XY$, $XZ$ and $YZ$ are antiparallel to $AB$, $AC$ and $BC$ with respect to the bisectors of triangle $ABC$. This implies that $\angle{BXC}=\angle{BYC}$ and hence that $\angle{AZC}=\angle{AXC}=\angle{AYB}=\angle{BZC}$. Hence $\angle{BZC}=\angle{AZC}=90^\circ$ which implies that $Z = H_1$. By the same argument, $Y=H_2$ and $X=H_3$. This completes the proof of the claim.

Dear Mathlinkers,
only a link
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=380345.
Sincerely
Jean-Louis

[quote="cadiTM"][quote="SnowEverywhere"]Now note that since $B'C' \| BC$  and $BC$ is tangent to the circumcircle of $A'B'C'$ at $T_1$, [/quote]

I think this part is wrong.. $T_1$ may not be on the circumcircle of $A'B'C'$ at this step.
But I think there is another way to solve this problem by that lemma...[/quote]

Sorry if I was unclear. I defined $A'B'C'$ to be the triangle inscribed in the incircle with sides parallel to those of $ABC$. By this definition, $T_1$ lies on the incircle and therefore the circumcircle of $A'B'C'$. I think it is right and have tried to make the definition clearer in the beginning.

Let $D,E,F$ be the points where the bisectors of $A$, $B$, and $C$ meet $BC$, $CA$, and $AB$, respectively. Let $U_1, U_2, U_3$ be the reflections of $T_1, T_2, T_3$ across $AD$, $BE$, and $CF$, respectively. We claim that the images of the reflections of $H_1H_2, H_2, H_3H_1$ across $T_1 T_2, T_2 T_3, T_3 T_1$ meet at $U_1, U_2, U_3$. To prove this claim, it suffices to show that the reflection of $U_1$ across $T_1 T_2$ lies on $H_1H_2$.

Denote by $d(\ell, X)$ the distance from $X$ to $\ell$, for any line $\ell$ and point $X$. Let $M$ be the midpoint of $T_1 T_2$. Let $H_1', H_2', U_1'$ be the images of the reflections of $H_1,H_2,U_1$ across $MC$. Because $MC \perp T_1 T_2$, $T_1 T_2$ is fixed by this reflection. Thus, it suffices to show that the reflection of $U_1'$ over $T_1 T_2$ lies on $H_1' H_2'$. Since $MC \perp T_1 T_2$ and $M$ is the intersection of $MC$ and $T_1T_2$, the reflection of $U_1$ across $MC$, reflected across $T_1 T_2$, is simply the reflection of $U_1$ across $M$. Therefore, we need only show that the reflection of $U_1$ across $M$, which we shall call $U_1''$, lies on $H_1'H_2'$. Since $H_1H_2$ is antiparallel to $AB$ with respect to $\angle C$, $H_1'H_2'$ is parallel to $AB$, so it is enough to show that $d(AB, U_1'') = d(AB, H_1'H_2')$.

Let $h_c = d(AB, C)$. We have $d(AB, H_1'H_2') = h_c - d(C, H_1'H_2') = h_c - d(C, H_1H_2) = h_c(1-\cos C)$, since $\triangle CH_1H_2$ and $\triangle CAB$ are similar with factor of similarity $\frac{H_2C}{CB} = \cos C$.

Since $U_1''$ is the reflection of $U_1$ across $M$, $d(AB, U_1'') = 2d(AB, M) - d(AB, U_1)$. We have $2d(AB, M) = d(AB,T_1) + d(AB, T_2) = (s-b) \sin B + (s-a) \sin A$. To compute $d(AB, U_1)$, we extend $DU_1$ to meet $AB$ at $C'$ and $AC$ at $B'$. $\triangle AB'C'$ is the reflection of $\triangle ABC$ across the bisector of $\angle A$, so $d(AB, U_1) = d(AC', U_1) = d(AC, T_1) = (s-c) \sin C$. Thus, if we denote by $K$ the area of $\triangle ABC$, we have

\begin{align*}
d(AB, U_1'')
&= 2d(AB, M) - d(AB, U_1) = (s-b) \sin B + (s-a) \sin A - (s-c) \sin C = (s-b) \frac{2K}{ac} + (s-a) \frac{2K}{bc} - (s-c) \frac{2K}{ab}\\
&= \frac{K}{abc} (b(a+c-b) + a(b+c-a) - c(a+b-c)) = \frac{2K}{c} \cdot \frac{2ab - a^2 - b^2 + c^2}{2ab} = h_c (1 - \cos C) = d(AB, H_1'H_2'),
\end{align*}
as desired.

Dear Mathlinkers,
you can see also

http://perso.orange.fr/jl.ayme    vol. 9 Le triangle réféchi

Sincerely
Jean-Louis

There is also a simple solution using complex numbers.  WLOG let $ \omega $, the incircle of $ \triangle{ABC}, $ be the unit circle and let the complex coordinates of $ A, B, C, T_1, T_2, T_3, H_1, H_2, H_3 $ be denoted by $ a, b, c, t_1, t_2, t_3, h_1, h_2, h_3 $ respectively.

Now, it is well-known that $ a = \frac{2t_{2}t_{3}}{t_2 + t_3}, b = \frac{2t_{1}t_{3}}{t_1 + t_3}, c = \frac{2t_{1}t_{2}}{t_1 + t_2} $.  Since $ H_2 $ is the projection of $ B $ onto "chord" $ T_{2}T_{2} $ of $ \omega $ we have that $ h_2 = \frac{1}{2}\left(b + 2t_2 - t_{2}^2\overline{b}\right) = \frac{t_{1}t_{2} + t_{2}t_{3} + t_{1}t_{3} - t_{2}^2}{t_1 + t_3} $.

Now, let $ P_2 $ be the reflection of $ H_2 $ over $ T_{2}T_{3} $ and denote its complex coordinate by $ p_2 $.  Then $ p_2 = t_2 + t_2 - t_{2}t_{3}\overline{h_2} = \frac{t_1(t_{2}^2 + t_{3}^2)}{t_2(t_1 + t_3)} $.  Letting $ P_3 $ be the reflection of $ H_3 $ over $ T_{2}T_{3} $ and denoting its complex coordinate by $ p_3 $ we similarly obtain that $ p_3 = \frac{t_1(t_{2}^2 + t_{3}^2)}{t_3(t_1 + t_2)} $.

Now let $ Z $ be an intersection of line $ P_{2}P_{3} $ with $ \omega $.  Let its complex coordinate be $ z $.  It is clear that $ z $ satisfies the following equation: $ \frac{z - p_2}{\overline{z} - \overline{p_2}} = \frac{p_2 - p_3}{\overline{p_2} - \overline{p_3}} $.

Now we can compute $ p_2 - p_3 = \frac{t_{1}^2(t_3 - t_2)(t_{2}^2 + t_{3}^2)}{t_{2}t_{3}(t_1 + t_3)(t_1 + t_2)} $ and that $ \overline{p_2} - \overline{p_3} = \frac{(t_2 - t_3)(t_{2}^2 + t_{3}^2)}{t_{2}t_{3}(t_1 + t_3)(t_1 + t_2)} $ so $ \frac{p_2 - p_3}{\overline{p_2} - \overline{p_3}} = -t_{1}^2 $.  Moreover, since $ Z \in \omega $ we have that $ \overline{z} = \frac{1}{z} $.

Therefore we have that $ \frac{z - p_2}{\frac{1}{z} - \overline{p_2}} = -t_{1}^2 \Longrightarrow z^2 - (p_2 + t_{1}^2\overline{p_2})z + t_{1}^2 = 0 $.  We can compute that $ p_2 + t_{1}^2\overline{p_2} = \frac{t_1(t_{2}^2 + t_{3}^2)}{t_{2}t_{3}} $.

By a quick application of the quadratic formula we find that the two possibilities for $ z $ are $ \frac{t_{1}t_{2}}{t_3} $ and $ \frac{t_{1}t_{3}}{t_2} $.  This immediately implies that the vertices of the triangle formed by the lines $ P_{1}P_{2}, P_{2}P_{3}, P_{3}P_{1} $ have complex coordinates $ \frac{t_{1}t_{2}}{t_3}, \frac{t_{2}t_{3}}{t_1}, \frac{t_{3}t_{1}}{t_2} $ all of which clearly lie on $ \omega $ as desired.

The motivation for using complex numbers is clear since, after taking the incircle to be the unit circle, all relevant points are projections or reflections with respect to chords of the circle, which are "nice" in complex coordinates.

Let $X_1$ be on the incircle such that $T_1X_1\parallel T_2T_3$, and define $X_2,X_3$ similarly. We claim that $X_1X_2X_3$ is the triangle formed. Note that
\[
\angle X_3T_1B=\angle T_3T_1T_2=\angle X_2T_1C
\]
so the sides are indeed parallel to the sides of $ABC$.

Let $T_1T_3$ meet $X_1T_2$ at $D$. By symmetry, $D$ lies on the angle bisector of $\angle BAC$, which is the perpendicular bisector of $T_3T_2$ and $T_1X_1$. So by a well known lemma, $D$ lies on the circle with diameter $AC$.

 Let $E$ be the reflection of $X_1$ over $T_1T_2$. It suffices to show that $E$ lies on $H_1H_3$.
\[
\angle DT_3T_2=\angle DT_2T_3=\angle CT_1T_2=\angle CT_2T_1
\]
so
\[
\angle H_1DA=\angle H_1CA=\angle T_1DX_1
\]
Thus $\angle X_1DA=\angle ADT_1=\angle T_1DH_1$, so $E$ is also the reflection of $T_1$ across $H_1D$. But $D$ is the midpoint of arc $H_3C$, so
\[
\angle DH_1E=\angle T_1H_1D=\angle CH_1D=180-\angle H_3H_1D
\]
and so $E$ lies on $H_1H_3$, as desired.

Let Ω be the homothety with negative radius taking the circumcircle to the inradius. Construct the triangle A'B'C' that is the image of ABC under Ω, and reflect each side w.r.t the respective side of TaTbTc. It is enough to prove the resulting triangle, call it HaHbHc, is the orthic triangle. But Ta is equidistant A'B' and A'C', and so circle centered at Ta is tangent to lines A'B', A'C', and the reflections of these across TaTb and TaTc. From this TaHa is parallel to BC and so Ha lies on BC. Similarly Hb and Hc lie on the sides of ABC and we finish easily.

My solution:
Let $I$ be the incenter of $\triangle{ABC}$, $AI, BI, CI\cap{BC, CA, AB} = D, E, F$, $X, Y, Z$ are the reflections of $T_1, T_2, T_3$ WRT $AD, BE, CF$
$\Rightarrow$ $DX, EY, FZ$ are tangent to $(I)$
$\Rightarrow$ $T_2T_3, EF, YZ$ are concurrent (familiar property of circumscribed quadrilateral) (1)
Moreover: $T_2T_3 = T_1Z$ (reflect WRT $CI$) and $T_2T_3 = T_1Y$ (reflect WRT $BI$) so $T_1Z = T_1Y \Rightarrow YZ\parallel{BC}$
Analogously, we have: $ZX\parallel{CA}, XY\parallel{AB}$
On the other hand:
$(EH_2AT_2)
= \frac{\overline{AE}}{\overline{AH_2}}:\frac{\overline{T_2E}}{\overline{T_2H_2}}
= \frac{\overline{AE}}{\overline{AH_2}}:\frac{\overline{IE}}{\overline{IB}}
= \frac{\overline{AE}}{\overline{AH_2}}:\frac{\overline{AE}}{\overline{AB}}
= \frac{\overline{AB}}{\overline{AH_2}}
= \frac{\overline{AC}}{\overline{AH_3}}
= \frac{\overline{AF}}{\overline{AH_3}}:\frac{\overline{AF}}{\overline{AC}}
= \frac{\overline{AF}}{\overline{AH_3}}:\frac{\overline{IF}}{\overline{IC}}
= \frac{\overline{AF}}{\overline{AH_3}}:\frac{\overline{T_3F}}{\overline{T_3H_3}}
= (FH_3AT_3)$
$\Rightarrow$ $EF, H_2H_3, T_2T_3$ concur (2)
(1), (2) $\Rightarrow$ $YZ, H_2H_3, T_2T_3$ concur (3)
Moreover: $(H_2H_3, T_2T_3) \equiv (T_2T_3, YZ)$ (mod $\pi$) (because $YZ\parallel{BC}$ and $H_2H_3, BC$ are antiparallel in $\angle{BAC}$, and $T_2T_3$ is perpendicular to the bisector of $\angle{BAC}$) (4)
(3), (4) $\Rightarrow$ $YZ$ is the reflection of $H_2H_3$ WRT $T_2T_3$
and the conclusion follows.
Q.E.D.

We use complex numbers with $\omega$ the unit circle. Let $T_1 = a$, $T_2 = b$, $T_3 = c$. The main content of the problem is to show that the triangle in question has vertices $ab/c$, $bc/a$, $ca/b$ (which is evident from a good diagram).

Since $A = \frac{2bc}{b+c}$, we have \[ H_1 = \frac{1}{2} \left( \frac{2bc}{b+c} + a + a - a^2 \cdot 	\frac{2bc}{b+c} \right) 	= \frac{ab+bc+ca-a^2}{b+c}. \] The reflection of $H_1$ over $\overline{T_1 T_2}$ is \begin{align*} 	H_1^C &= a + b - ab \overline{H_1} \\ 	&= a + b - b \cdot \frac{ac+ab+a^2-bc}{a(b+c)} \\ 	&= \frac{a(a+b)(b+c) - b(a^2+ab+ac-bc)}{a(b+c)} \\ 	&= \frac{c(a^2+b^2)}{a(b+c)}. \end{align*} Now, we claim that $H_1^C$ lies on the chord joining $\frac{ca}{b}$ and $\frac{cb}{a}$, which implies the problem (since by analogy $H_1^B$ does as well). To see this, it suffices to compute \begin{align*} 	H_1^C + \left( \frac{ca}{b} \right)\left( \frac{ca}{a} \right) \overline{H_1^C} 	&= \frac{c(a^2+b^2)}{a(b+c)} 		+ c^2 \frac{\frac 1c \cdot \frac{a^2+b^2}{a^2b^2}} 		{\frac1a\left( \frac{b+c}{bc} \right)} \\ 	&= \frac{c(a^2+b^2)}{a(b+c)} 		+ \frac{c(a^2+b^2)}{abc^{-1}(b+c)} \\ 	&= \frac{c(a^2+b^2)}{a(b+c)} \left( \frac{b+c}{b} \right) \\ 	&= \frac{c(a^2+b^2)}{ab}= \frac{ca}{b} + \frac{cb}{a} \end{align*} as desired.

Easy to see that the the symmetric images of  $ H_2H_3$ wrt the lines $ T_2T_3$ ( called it $ l_1$) is parallel to $BC$
Let the Medial triangle of $ABC$ be $DEF$
Consider the Feuerbach hyperbola $f$ of $ABC$, Feuerbach point of $ABC$ be $V$, $H_2H_3$ cut $T_2T_3$ at $K_1$.
Then $K_1$ is the pole of $BC$ wrt $f$, so $l_1$ is the polar of $D$ wrt $f$.
So it's sufficient to prove that the pole of $EF$ wrt $f$ is on the incircle of $ABC$.
Let $EF$ cut $T_2T_3$ at $P_1$. From Fontene theorem, $D,P_1,V$ are collinear.
Note that the pole of $V$ pass through $T_1$ and parallel to $T_2T_3$, and the pole of the infinity point of $BC$ is $VD$, and they meet on the incircle of $ABC$ (by inversion of $D$), so the problem is done.

Nice problem with a very rich configuration :)

Let $X_i$ be the point on the incircle $\omega$ so that $TX_i\parallel X_{i+1}X_{i+2}$. We claim that $X_1X_2X_3$ is the desired triangle. In particular, we will show that $X_iX_{i+1}$ is the reflection of $H_iH_{i+1}$ over $T_iT_{i+1}$, which will finish. Actually, we'll just show that $X_1$ is on the reflection of $H_1H_2$ over $T_1T_2$; the rest follows similarly. If $AB = AC$ the result is immediate from $H_1 = T_1 = X_1$, so assume now that $AB < AC$.

If $\ell$ be the reflection of $H_1H_2$ over $T_1T_2$, then since $H_1H_2$ and $AB$ are antiparallel and $T_1T_2$ is perpendicular to an angle bisector of $\angle ACB$, we get $\ell \parallel AB$. Let $\ell'$ be the line through $X_1$ parallel to $\ell\parallel AB$, then we wish to show $\ell = \ell'$. Let $\ell$ meet $AI$ at $P$, where $I$ is the incenter of $ABC$ and $D = AI\cap BC$. We note that since $T_1P, X_1P$ are reflections over $AI$ (from $T_1X_1\perp AI$) we obtain $T_1P\parallel AC$ since $AC, AB$ are reflections over $AI$ and $AB\parallel X_1P$.

Thus, $\triangle DTP \sim \triangle DCA$. Now, \[
\frac{DT_1}{H_1T_1} = \frac{DI}{IA} = \frac{CD}{DA} = \frac{DT_1}{T_1P},
\]where the equalities follow from $T_1I\parallel AH_1$, the Angle Bisector Theorem from $CI$ bisecting $\angle ACD$, and $\triangle DTP \sim \triangle DCA$. Thus $H_1T_1 = T_1P$. Note that since $AB < AC$, $A, B, H_1$ are on the same side of $T_1I$, so $CH_1 > CT_1$. Thus, the reflection of $H_1$ over $T_1T_2$ lies on the line through $T_1$ parallel to $AC$, lies the same distance from $T_1$ as $H_1$, and is on the opposite side of $BC$ as the incircle.

Now, since $T_1, D, C$ lie on $BC$ in that order, it follows that $\frac{AD}{DP} = \frac{T_1D}{DC}$ where we direct segments, by considering the homothety centered at $D$ taking $T_1P\to CA$. Thus, $A,P$ lie on opposite sides of $BC$, so $P$ is the reflection of $H_1$ over $T_1T_2$. Now, the line through $P$ parallel to $AB$ is $\ell$, so $\ell = \ell'$ and $X_1\in \ell$ as desired. Similarly, $X_1$ is on the reflection of $H_1H_3$ over $T_1T_3$, and so $X_1$ is indeed one of the vertices of our triangle, as are $X_2, X_3$, which completes the proof. $\blacksquare$

I think this solution is new.

Let $I,O$ be the incenter and circumcenter of $\triangle ABC$. First, since $H_2H_3\perp AO$, and the reflection of $AO$ over $T_2T_3$ gives a line perpendicular to $BC$, it follows that the reflection of $H_2H_3$ is parallel to $BC$. Now let $A_1$ be the incenter of $\triangle AH_2H_3$. By a well-known lemma, $A_1$ is also the orthocenter of $AT_2T_3$ (indeed, this is true because $AA_1 = AI\cos A$).

Now when we reflect triangle $A_1H_2H_3$ about $T_2T_3$, our previous observation tells us that the image of $A_1$ is $I$, so we get some triangle $IXY$, where $XY||BC$ as noted earlier. Furthermore, the distance from $I$ to $XY$ is the distance from $A_1$ to $H_2H_3$, which is just $r\cos A$ (since $AH_2H_3, ABC$ are similar with ratio $\cos A$). Therefore, the three reflected lines in the problem statement determine some triangle $A_2B_2C_2$ which is homothetic and therefore similar to $ABC$, and which satisfies that there exists a point $I$ in the plane whose distances to the three sides of the triangle are $r\cos A, r\cos B, r\cos C$. The only point $P$ in the plane such that the ratios of its projections to the sides of $A_2B_2C_2$ are $\cos A:\cos B:\cos C$ is the circumcenter of $A_2B_2C_2$, so $I$ is the circumcenter and $A_2B_2C_2$ has circumradius $r$, meaning it lies on the incircle, and we're done.

(There are some configuration issues, but since $ABC$ is acute, we can verify that $A_1$ lies on the "correct side" of $H_2H_3$, so $I$ lies on the "correct side" of $B_2C_2$ and is therefore in the interior of $A_2B_2C_2$, so this is fine.)

Let $Y$ lie on the incircle such that $\overline{T_2Y}\parallel\overline{T_3T_1}$. We will show that $Y$ lies on the reflection of $\overline{H_2H_3}$ over $\overline{T_2T_3}$, which is sufficient by symmetry.

    [asy]
        size(9cm);
        defaultpen(fontsize(10pt));

        pen pri=springgreen;
        pen sec=red;
        pen tri=blue;
        pen qua=purple;
        pen qui=orange;
        pen fil=invisible;
        pen sfil=invisible;
        pen tfil=invisible;
        pen qfil=invisible;
        pen qifil=invisible;

        pair A, B, C, H1, H2, H3, I, T1, T2, T3, X, Y, Z, P, Yp;
        A=dir(110);
        B=dir(200);
        C=dir(340);
        H1=foot(A, B, C);
        H2=foot(B, C, A);
        H3=foot(C, A, B);
        I=incenter(A, B, C);
        T1=foot(I, B, C);
        T2=foot(I, C, A);
        T3=foot(I, A, B);
        X=2*foot(T2+T3-T1, T2, T3)-(T2+T3-T1);
        Y=2*foot(T3+T1-T2, T3, T1)-(T3+T1-T2);
        Z=2*foot(T1+T2-T3, T1, T2)-(T1+T2-T3);
        P=extension(T1, Y, T2, T3);
        Yp=2*foot(Y, T2, T3)-Y;

        filldraw(A -- B -- C -- A -- cycle, fil, pri);
        filldraw(incircle(A, B, C), fil, pri);
        filldraw(T1 -- T2 -- T3 -- cycle, sfil, sec);
        filldraw(X -- Y -- Z -- cycle, tfil, tri);
        draw(T1 -- P -- T2, qua);
        draw(H1 -- H2 -- H3 -- H1, qui+dotted);
        draw(H2 -- Yp, qui+dotted);
        draw(Y -- Yp, tri+dashed);
        draw(H2 -- P, qui+dotted);
        draw(arc((B+C)/2, C, B), qui);
        fill(arc((B+C)/2, C, B) -- cycle, qifil);
        draw(B -- P, qui+dashed);

        dot("$X$", X, S);
        dot("$Y$", Y, SE);
        dot("$Z$", Z, dir(195));
        dot("$Y'$", Yp, N);
        dot("$A$", A, N);
        dot("$B$", B, SW);
        dot("$C$", C, SE);
        dot("$T_1$", T1, S);
        dot("$T_2$", T2, N/2);
        dot("$T_3$", T3, dir(150));
        dot("$H_1$", H1, S);
        dot("$H_2$", H2, N);
        dot("$H_3$", H3, dir(120));
        dot("$P$", P, E);
    [/asy]

Let $P=\overline{T_1Y}\cap\overline{T_2T_3}$, which lies on $\overline{BI}$ by reflection, and let $Y'$ be the reflection of $Y$ over $\overline{T_2T_3}$. Then by the Iran Lemma, $P$ lies on $(BCH_2H_3)$. Since $\measuredangle H_2PB=\measuredangle H_2CB=\measuredangle T_2PY$, but $\overline{PB}$ bisects $\angle T_2PY$, $\overline{PH_2}$ bisects $\angle T_2PY'$, so we deduce by $PY'=PY=PT_2$ that $PT_2H_2Y'$ is a kite. Now \[\measuredangle PH_2Y'=\measuredangle T_2H_2P=\measuredangle CBP=\measuredangle PBH_3=\measuredangle PH_2H_3,\]
completing the proof.

Pretty long but straightfoward by complex number.
Anw first G8 Ive solved :)
Set all of these at complex plane. Let the incircle touchpoint of $\triangle ABC$ opposite to $A,B,C$ be $a,b,c$ respectively in complex number. Therefore, $A = \frac{2bc}{b+c}, B = \frac{2ac}{a + c}, C = \frac{2ab}{a+b}$. \\
Since $H_1$ is the projection of $A$ towards $BC$, which gives us
\[ H_1 = \frac{ab + ac + bc - a^2}{b + c} \]
Similarly, $H_2 = \frac{ab + ac + bc - b^2}{a +c} $ and $H_3 = \frac{ab + ac + bc - c^2}{a+b}$\\
Now, we'll find the reflection of $H_1H_2$ over $T_1T_2$. \\
Reflection of $H_1$ across $T_1T_2$ is
\begin{align*}
& a + b - ab \cdot \frac{\frac{1}{ab} + \frac{1}{ac} + \frac{1}{bc} - \frac{1}{a^2}}{ \frac{1}{b} + \frac{1}{c} } \\  &= a + b - \frac{b(ac + ab + a^2 - bc)}{a(b+c)} \\
 &= \frac{a(a+b)(b+c) - b(ac + ab + a^2 - bc)}{a(b+c)} \\
 &= \frac{c(a^2 + b^2)}{a(b+c)}
\end{align*}
Similarly, we will get the other point as $ \frac{c(a^2 + b^2)}{b(a+c)} $
\\
Therefore, the three lines are determined by lines passing through \[ \left( \frac{c(a^2 + b^2)}{a(b+c)} , \frac{c(a^2 + b^2)}{b(a+c)} \right), \left( \frac{a(b^2 + c^2)}{b(a+c)} , \frac{a(b^2 + c^2)}{c(a+b)} \right), \left( \frac{b(a^2 + c^2)}{a(b+c)}, \frac{b(a^2 + c^2)}{c(a+b)} \right) \]
Now, the conjugate are
\[ \left( \frac{a^2 + b^2}{ab(b+c)}, \frac{a^2 +b^2}{ab(a+c)} \right),  \left( \frac{b^2+c^2}{bc(a+c)}, \frac{b^2+c^2}{bc(a + b)} \right) , \left( \frac{a^2 + c^2}{ac(a+b)} , \frac{a^2 + c^2}{ac(b+c)} \right) \]
Now, we want to determine the intersection of these three lines using the complex intersection formula. It suffices to find for one point as the others are analogously symmetrical.
\\
\begin{align*}
\text{Numerator} &= \frac{a^2c(a^2 + b^2)^2 (b - a)(c - b)(b^2 + c^2)}{a^2 b^3c (a+c)(b+c)(a+b)(a+c)} - \frac{c^2a(a^2 + b^2)(b-a)(b^2+c^2)^2(c-b)}{ab^3 c^2 (a+c)^2 (b+c)(a+b)} \\
 &= \frac{(b-a)(a^2 + b^2)^2 (c - b)(b^2 + c^2)}{b^3(a + c)^2(a+b)(b+c)} - \frac{(b^2+c^2)^2(c-b)(b-a)(a^2 + b^2)}{b^3(a+c)^2(b+c)(a+b)} \\
 &= \frac{(c-b)(b-a)(a^2 + b^2)(b^2+c^2)(a^2 - c^2)}{b^3(a + c)^2 (b+c)(a+b)} \\
 &= \frac{(c-b)(b-a)(a-c)(a^2+b^2)(b^2+c^2)}{b^3(a+c)(b+c)(a+b)}
 \end{align*}
 and furthermore, we have
 \begin{align*}
 \text{Denominator} &= a^2 (b^2 + c^2) \left( \frac{c-b}{bc(a+c)(a+b)} \right)  \left( \frac{b^2 + c^2}{bc} \right)  \frac{a - b}{(a+c)(b+c)} -  c^2 (a^2 + b^2) \left( \frac{b - a}{ab(a+c)(b+c)} \right)  \left( \frac{b^2+c^2}{bc} \right)  \frac{b-c}{(a+b)(a+c)} \\
 &= \frac{(a^2 + b^2)(b^2+c^2)(a-b)(c-b)a^2}{ab^2c(a+c)^2(a+b)(b+c)} - \frac{c^2(a^2 + b^2)(b^2+c^2)(a-b)(c-b)}{ab^2c (a+c)^2(a+b)(b+c)} \\
 &= \frac{(a^2+b^2)(b^2+c^2)(a-b)(b-c)(c-a)}{ab^2c(a+c)(a+b)(b+c)}
 \end{align*}
 Divide the two of them, we get the intersection being
 \[ \frac{\frac{(c-b)(b-a)(a-c)(a^2+b^2)(b^2+c^2)}{b^3(a+c)(b+c)(a+b)} }{\frac{(a^2+b^2)(b^2+c^2)(a-b)(b-c)(c-a)}{ab^2c(a+c)(a+b)(b+c)} } = \frac{-ac}{b} \]
 Notice that $-\frac{ac}{b}$ lies on the unit circle since
 \[ -\frac{\overline{ac}}{\overline{b}} = -\frac{\frac{1}{ac}}{\frac{1}{b}} = -\frac{b}{ac} = \frac{1}{-\frac{ac}{b}}\]

Pretty hard.

[hide=Complex]
We use complex numbers. Let the incircle be the unit circle. Let $T_1=x,T_2=y,T_3=z$. Then
\[ a=\frac{2yz}{y+z}, \ b = \frac{2xz}{x+z}, \ c = \frac{2xy}{x+y}.  \]
Now, $H_1$ is the foot from $A$ to $T_1T_1$, which is
\[ h_1 = \frac12 (2x+a-x^2 \overline{a}) = x + \frac{yz}{y+z} - \frac{x^2}{y+z} = \frac{xy+yz+zx - x^2}{y+z}.\]
Similarly,
\[ h_2 = \frac{xy+yz+zx - y^2}{x+z}, \ h_3 = \frac{xy+yz+zx - z^2}{x+y}.  \]
We want to reflect $H_2H_3$ over $T_2T_3$. To do this, we reflect $H_2$ and $H_3$ over $T_2T_3$ individually, and then consider the new line through $H_2'$ and $H_3'$. We have
\begin{align*}
h_2' &= y+z - yz \overline{h_2} \\
&= y+z - yz \cdot \frac{\frac{1}{xy}+\frac{1}{yz} + \frac{1}{zx} - \frac{1}{y^2}}{\frac{1}{x} + \frac{1}{z}} \\
&= y+z - yz\cdot \frac{x+y+z-\frac{xz}{y}}{y(x+z)} \\
&= y+z - \frac{z(xy+y^2+yz-xz)}{y(x+z)} \\
&= \frac{y(y+z)(x+z) - z(xy+y^2+yz-xz)}{y(x+z)} \\
&= \frac{x(y^2+z^2)}{y(x+z)} \\
\implies &h_2' = \frac{x(y^2+z^2)}{y(x+z)}, \ h_3' = \frac{x(y^2+z^2)}{z(x+y)}.
\end{align*}
We want to find where $H_2'H_3'$ intersects the unit circle, call this point $D$. We know $d=kh_2'+(1-k)h_3'$ for some $k \in \mathbb{R}$, which is
\begin{align*}
    d &= kh_2'+(1-k)h_3' = h_3'+(h_2'-h_3')k = \frac{x(y^2+z^2)}{z(x+y)} + x(y^2+z^2) \left(\frac{1}{y(x+z)} - \frac{1}{z(x+y)}\right) k \\
    &= \frac{x(y^2+z^2)}{z(x+y)} + \frac{x^2(y^2+z^2)(z-y)}{yz(x+y)(x+z)}k.
\end{align*}
Then
\[ \overline{d} = \frac{x(y^2+z^2)}{xyz(x+y)} + \frac{(y^2+z^2)(y-z)}{yz(x+y)(x+z)}k,\]
We also have $d\overline{d}=1$, so
\begin{align*}
    &d\overline{d} = \left(\frac{x(y^2+z^2)}{z(x+y)} + \frac{x^2(y^2+z^2)(z-y)}{yz(x+y)(x+z)}k\right)\left(\frac{x(y^2+z^2)}{xyz(x+y)} + \frac{(y^2+z^2)(y-z)}{yz(x+y)(x+z)}k \right)=1 \\
    \implies& \frac{-x^2(y^2+z^2)^2(y-z)^2}{y^2z^2(x+y)^2(x+z)^2}k^2 + \frac{x(y^2+z^2)^2(y-z)(y-x)}{y^2z^2(x+y)^2(x+z)}k + \frac{(xy-z^2)(y^3-xz^2)}{yz^2(x+y)^2} = 0 \\
    \implies& -x^2(y^2+z^2)^2(y-z)^2 k^2 + x(y^2+z^2)^2(y-z)(y-x)(x+z) k + y(x+z)^2(xy-z^2)(y^3-xz^2)=0.
\end{align*}
It turns out that the discriminant of this quadratic factors as
\[ x^2(y^2+z^2)^2(y-z)^4(x+z)^2(x+y)^2(y+z)^2.\]
Solving the quadratic yields one solution to be
\[ k=\frac{y(x+z)(xy-z^2)}{x(y^2+z^2)(z-y)}. \]
Plugging this back in to get $d$ yields
\begin{align*}
    d &=\frac{x(y^2+z^2)}{z(x+y)} + \frac{x^2(y^2+z^2)(z-y)}{yz(x+y)(x+z)}\cdot \frac{y(x+z)(xy-z^2)}{x(y^2+z^2)(z-y)} \\
    &= \frac{x(y^2+z^2)}{z(x+y)} + \frac{x(xy-z^2)}{z(x+y)} \\
    &= \frac{xy^2+x^2y}{z(x+y)} \\
    &= \frac{xy}{z}.
\end{align*}
By symmetry of the $x,y$ in the numerator, this must also be the intersection of another one of the lines with the unit circle.
[/hide]

Another complex numbers solution, I'm posting because it uses a slightly different approach.

Let $P, Q, R$ be the points on $\omega$ such that $T_1$ is the midpoint of arc $RPQ$, and symmetric results hold.
It suffices to prove that reflection of $R$ in $T_2T_3$ lies on $H_2H_3$, since then other symmetric results hold.

We use complex numbers with $\omega$ as unit circle. We may let $p = x^2, q = y^2, r = z^2, t_1 = yz, t_2 = zx, t_3 = xy$.
Then by the intersection of tangents formula, $a = 2xyz/(y+z)$, and other symmetric results hold.
By foot of perpendicular formula (on chord $T_3T_3$), $h_3 = xyz/(x+y) + xy - x^2y^2/((x+y)z)$, and other symmetric results hold.
Let $R'$ be the reflection of $R$ in $T_2T_3$, then by reflection formula, $r' = xy + xz - x^2y/z$.
Now, it suffices to prove that $(r'-h_2)/(r'-h_3)$ is real.
The expression simplifies to $y(xy-z^2)(x+z)/((xy^2-z^3)(x+y))$, which is equal to its conjugate, so we're done.

here is my solution for this nice problem
 (we assume WLOG $AB\ge BC\ge AC$)
let $C'$ be the reflection of $C$ over $T_1T_2$, we have $CT_1=CT_2$ so $CT_1C'T_2$ must be a rhombus, hence $T_1H'_1$ is parallel to $AC$
Let $X=(T_1T_2) \cap (AI)$ (this point lies on the circle with diameter $AB$)
We have $\widehat{T_1XA}=\widehat{T_2XA}=\frac{\widehat{B}}2$
and $\widehat{T_1XH_1}=\widehat{H_1XA}-\widehat{T_2XA}=\widehat{B}-\frac{\widehat{B}}2$
so $H'_1$ lies on $(AI)$ ( the case where $H_1$ is the one that lies on $(AI)$ gives also $H'_1$ lies on $(AI)$ because in that case $H'_1=H_1$)
by symmetry we can prove the same claim for the other points.
We also have $\widehat{IH'_1H'_2}=\widehat{IH'_1T_1}-\widehat{H'_2H'_1T_1}=180-\frac{\widehat{A}}2-\widehat{T_1H_1H_2}=180-\frac{\widehat{A}}2-(180-\widehat{A})=\frac{\widehat{A}}2$ so $H'_1H'_2$ is parallel to $AB$
So $H'_1$ and $H''_1$ lie on $AI$ with $H''_1$ the reflection of $H_1$ over $T_1T_3$

I claim that the intersection of $H'1H'2$ and $H''_3H''_1$ is the reflection of $T_1$ over $AI$ $Y$ , this claim clearly solves the problem because the reflection of $T_1$ over $AI$ lies on the incircle since $T_2$ and $T_3$ and $I$ are fixed by this reflcection and so is the incircle
We have $\widehat{YH'_1A}=\widehat{T_1H'_1A}=180-\frac{\widehat{A}}2$ so it lies on $H'_1H'_2$ by symmetry we prove that it lies on $H''_1H''_3$
QED

We use complex numbers. Set $\omega$ as the unit circle, and let $t_1, t_2, t_3$ be the free variables. From the tangent intersection formula, we have $a=\frac{2t_Bt_C}{t_B+t_C}$, $b=\frac{2t_At_C}{t_A+t_C}$, and $c=\frac{2t_At_B}{t_A+t_B}$. Next, using the foot of altitude formula, we have $h_A=\frac{-t_A^2+\sum_{cyc} t_At_B}{t_B+t_C}$, and $h_B=\frac{-t_C^2+\sum_{cyc} t_At_B}{t_A+t_C}$. Now, let $Z_A$ be the reflection of $H_A$ over $T_AT_B$, and $Z_B$ be the reflection of $H_B$ over $T_AT_B$. Using the complex reflection formula, we find $z_A=t_A+t_B-t_At_B\bar{h_A}=\frac{t_C(t_A^2+t_B^2)}{t_A(t_B+t_C)}$, and $z_B=\frac{t_c(t_A^2+t_B^2)}{t_B(t_A+t_C)}$.

Now we compute the intersections of $Z_AZ_B$ with $\omega$. We have the equation $x(\overline{z_A-Z_B})+\bar{x}(z_B-z_A)+(z_A\bar{z_B}-\bar{z_A}z_B)=0$, which, since $|x|=1$, becomes $x^2(\overline{z_A-z_B})+(z_A\bar{z_B}-\bar{z_A}z_B)x+(z_B-z_A)=0$. We can compute $\overline{z_A-z_B}=\frac{(t_A^2+t_B^2)(t_A-t_B)}{t_At_B(t_B+t_C)(t_A+t_C)}$, $z_A\bar{z_B}-\bar{z_Az_B}=\frac{(t_A^2+t_B^2)^2t_C(t_A-t_B)}{t_A^2t_B^2(t_B+t_C)(t_A+t_C)}$, and $z_B-z_A=\frac{(t_A-t_B)t_C^2(t_A^2+t_B^2)}{t_At_B(t_B+t_C)(t_A+t_B)}$. Thus, after dividing out common factors and multiplying by $t_At_B$, the quadratic becomes $t_At_Bx^2+(t_A^2+t_B^2)t_Cx+t_At_Bt_C^2=0$. We can now apply quadratic formula to get $x=-\frac{t_Bt_C}{t_A}, -\frac{t_At_C}{t_B}$, so $l_C\cap \omega=-\frac{t_Bt_C}{t_A}, -\frac{t_At_C}{t_B}$. By symmetry (swapping variables), we get $l_B\cap \omega=-\frac{t_At_B}{t_C}, -\frac{t_Bt_C}{t_A}$, and $l_A\cap \omega=-\frac{t_At_B}{t_C}, -\frac{t_At_C}{t_B}$. Therefore, the three vertices of the triangle formed by $l_A, l_B, l_C$ are $-\frac{t_At_B}{t_C}, -\frac{t_At_C}{t_B}, -\frac{t_Bt_C}{t_A}$, which are all on the unit circle, so we are done.

[quote=mathlogician]@above is this how you treat a very very nice problem :([/quote]

You gotta do what you gotta do :D

Nothing new here.

Let $DEF$ be the triangle determined by the three lines. A well-drawn diagram will inspire us to guess that $DT_1 \parallel T_2T_3$ and likewise equivalents. Thus it is sufficient (bearing in mind symmetry) to show that the reflection of $H_1$ and $H_2$ over $T_1T_2$ lie on the chord $DE$, where we redefine $D$ and $E$ as the points of intersection of the line through $T_1$ parallel to $T_2T_3$ and the line through $T_2$ parallel to $T_3T_1$, respectively.

We use complex numbers with the incircle of $\triangle ABC$ as the unit circle. As usual, let lowercase letters denote the complex coordinates of the point represented by their respective uppercase letter unless otherwise stated. For convenience's sake, let $t_1 = x, t_2 = y, t_3 = z$.

Then
\[ a = \frac{2yz}{y+z}, b = \frac{2zx}{z+x}, c = \frac{2xy}{x+y}. \]
Because $H_1$ is the foot from $A$ to $T_1T_1$, we obtain
\[ h_1 = \frac{1}{2}( 2x - a + x^2 \overline{a}) = \frac{xy + yz + zx - x^2}{y+z} . \]
------
Let $H_1'$ denote the reflection of $H_1$ over $T_1T_2$.
\begin{align*}
h_{1}' =& x + y - xy \overline{h_1} \\
&= x + y - \frac{xy(x(x+y+z) - yz)}{x^2(y+z)} \\
&= \frac{ x(x+y)(y+z) - xy(x+y+z) + y^2z}{x(y+z)} \\
&= \frac{x^2(y+z) - x^2y + y^2z}{x(y+z)} = \frac{z(x^2 + y^2)}{x(y+z)}
\end{align*}
Moreover, a straightforward computation gives $\overline{h_1'} = \frac{x^2 + y^2}{xy(y+z)}$.
------
Now we check that $H_{1}' \in DE$. It suffices to show taht
\[ h_1' + de \overline{h_{1}'} = d + e . \]
Since $DT_1 \parallel T_2T_3$, we get $d = \frac{yz}{x}$. Likewise, $e = \frac{zx}{y}$. Thus
\begin{align*}
 h_1' + de \overline{h_1'} &= \frac{z(x^2+y^2)}{x(y+z)} + \frac{z^2(x^2+y^2)}{xy(y+z)} \\
&= \frac{ z(x^2+y^2)(y+z))}{xy(y+z)} \\
&= \frac{z(x^2+y^2)}{xy} = \frac{zx}{y} + \frac{zy}{x} \\
&= d + e,
\end{align*}
as desired. By symmetry, the reflection of $H_2$ over $T_1T_2$ also lies on chord $DE$, so we are done.

The key observation is that the triangle is formed from isosceles-trapezoiding the intouch. More specifically, $T_1T_2 [\text{insertvertex}] T_3$ is an isosceles trapezoid, and this is cyclic.

Set $T_1T_2T_3$ unit circle, at $x$, $y$, and $z$ respectively because subscripts are hard to type. Let DEF be the vertices, such that $T_1DT_2T_3$ is an isosceles trapezoid, etc. Then by isosceles trapezoid trick, $d = \frac{yz}{x}$, cyclic.

Now, we need to calculate the reflections of the feet of altitudes. $a = \frac{2yz}{y + z}$ by ice cream cone; then foot of $a$ onto $xx$ gives $h_1 = x + \frac{yz - x^2}{y + z}$, which has conjugate $\frac 1 x + \frac{x^2 - yz}{x^2(y + z)}$. Now, reflect over $xy$; we get $h_1' = x - \frac{y(x^2 - yz)}{x(y + z)} = \frac{z(x^2 + y^2)}{x(y + z)}$.

Now, determinants. We have \begin{align*}
    \begin{vmatrix} \frac{yz}{x} & \frac{x}{yz} & 1 \\ \frac{xz}{y} & \frac{y}{xz} & 1 \\ \frac{z(x^2 + y^2)}{x(y + z)} & \frac{x^2 + y^2}{xy(y + z)} & 1 \end{vmatrix} &\propto \begin{vmatrix} \frac y x & \frac{x}{yz} & 1 \\ \frac x y & \frac{y}{xz} & 1 \\ \frac{x^2 + y^2}{x(y + z)} & \frac{x^2 + y^2}{xy(y + z)} & 1 \end{vmatrix} \\
    &= \frac{x^2 + y^2}{xy(y + z)} \cdot \left(\frac{x}{y} - \frac{y^2}{xz} - \frac{y}{x} + \frac{x}{z} \right) + \frac{y}{xz} \cdot \left( \frac y x - \frac x z \right) - \frac{x}{yz} \left( \frac x y - \frac{y^2}{xz} \right) \\
    &= \frac{x^2 + y^2}{y^2z} - \frac{x^2 + y^2}{x^2z} + \frac{y^2}{x^2z} - \frac{x^2}{y^2z} \\
    &= 0,
\end{align*} as desired. The others are similar, so we are done.

Let $I$ be the incenter of $\triangle ABC$, and $D$, $E$ the reflections of $H_1$ over $T_3T_1$ and $T_1T_2$, respectively. Let $K$ be the meeting point of the reflections of $H_3H_1$ over $T_3T_1$ and $H_1H_2$ over $T_1T_2$. Then by symmetry it suffices to prove that $KT_1T_2T_3$ is cyclic.

[hide=Diagram]
[img]https://cdn.discordapp.com/attachments/756987494393708665/870014346225405982/unknown.png[/img]
[/hide]

[b]Lemma 1: $D$, $E$ lie on $AI$.[/b]

[b]Proof:[/b] Let $X$, $Y$, $M$ be the midpoints of $AH_1$, $IH_1$, $AI$, respectively and let $Z$ be the foot of $H_1$ to $T_3T_1$. Then let the perpendicular bisectors of $AH_1$ and $IH_1$ meet $T_3T_1$ at $P$, $Q$ respectively. Let $O = XP \cap YQ$ and $L = AI \cap T_3T_1$.

By the Iran Lemma, we see that $\angle CLA = 90^\circ$. Since $X$ is the midpoint of $AH_1$ and $OX \perp AH_1$, $OX$ passes through the midpoints of $AB$ and $CA$ and so again by the Iran Lemma we see that $P$ lies on $CI$ and that $\angle CPA = 90^\circ$.

Thus $CLH_1PA$ is cyclic. We have $\angle H_1OY = \angle IOY = \angle IMY = \angle LAH_1 = \angle QPH_1$, so $OPQH_1$ is cyclic. Then by Simson Line from $H_1$, we see that $X$, $Y$, $Z$ are collinear and by a homothety of scale $2$ from $H_1$, $D$ lies on $AI$. Similarly, we can find that $E$ also lies on $AI$.

[b]Lemma 2: $KD \parallel ET_1$ and $KE \parallel DT_1$[/b]

[b]Proof:[/b] Define $F = KD \cap CI$, $G = T_3T_1 \cap KF$, and $S = H_1H_3 \cap CI$.
$H_1E \parallel CI$ so $\angle T_1EH_1 = \angle T_1CI = \frac{\angle C}{2}$. We have:
$$\angle KFC + \angle FGS + \angle ICH_1 = \angle H_1SC + \angle SCH_1 = \angle H_3H_1B = \angle A.$$
Notice that:
$$\angle FGS = 2\angle H_1GT_1 = 2(\angle H_3H_1B - \angle T_3T_1B) = \angle A - \angle C.$$
Then we have:
$$\angle KFC + (\angle A - \angle C) + \frac{\angle C}{2} = \angle A \Rightarrow$$
$$\angle KFC = \angle T_1EH_1 = \frac{\angle C}{2}.$$
Thus, $KD \parallel ET_1$ and similarly $KE \parallel DT_1$.

$KDT_1E$ is a parallelogram, but $T_1D = T_1H_1 = T_1E$, so $KDT_1E$ is also a rhombus. Then $AI$ is the perpendicular bisector of $KT_1$ and of $T_2T_3$, so $KT_1T_2T_3$ is cyclic as desired. $\square$

12 hours :stretcher:

[asy]
size(250);
pair T1 = dir(47.26-137.26), T2 = dir(47.26), T3 = dir(47.26+119.74);
pair A = (2*T2*T3)/(T2+T3), B = (2*T1*T3)/(T1+T3), C = (2*T1*T2)/(T1+T2);
pair D = (T2*T3)/T1, EE = (T1*T3)/T2, F = (T1*T2)/T3;
pair H1 = foot(A, B, C), H2 = foot(B, A, C), H3 = foot(C, A, B);
pair X1 = extension(H2, H3, T2, T3), X2 = extension(H1, H3, T1, T3), X3 = extension(H1, H2, T1, T2);
pair Y = extension(T2, D, T3, T1);
pair D1 = 2*foot(D, T1, Y)-D;
draw(A--B--C--cycle, fuchsia);
draw(incircle(A, B, C), red);
draw(circumcircle(A, H1, H3), magenta);
draw(circumcircle(C, Y, T1), yellow);
draw(A--Y--C, brown);
draw(X2--D, orange);
draw(X2--Y, heavyred);
draw(X2--D1, orange);
draw(H2--H3, orange+dashed);
draw(T1--T3, heavyred+dashed);
draw(EE--F, orange+dashed);
draw(H1--X3, orange+dashed);
draw(T1--X3, heavyred+dashed);
draw(D--X3, orange+dashed);
draw(T1--D1, red+dashed);
draw(T3--T2--Y, brown);
draw(T1--D, brown);
draw(D1--Y, orange);
draw(rightanglemark(A, Y, C, 4), brown);
dot("$A$", A, N);
dot("$B$", B, SW);
dot("$C$", C, SE);
dot("$D'$", D, SE);
dot("$E'$", EE, NE);
dot("$F'$", F, NW);
dot("$H_1$", H1, S);
dot("$H_2$", H2, N);
dot("$H_3$", H3, NW);
dot("$T_1$", T1, S);
dot("$T_2$", T2, NE);
dot("$T_3$", T3, NW);
dot("$Y$", Y, SW);
dot("$X_1$", X1, dir(60));
dot("$X_2$", X2, NW);
dot("$X_3$", X3, NE);
dot("$P$", D1, SW);
[/asy]

WLOG $AC>AB$ (to avoid annoying configuration issues). Denote by $D', E', F'$ points on the incircle $\omega$ that satisfy $\overline{T_1D'} \parallel \overline{T_2T_3}$ and cyclic permutations. We will show that $D' = D, E' = E$ and $F'=F$. Notice that it suffices to show that the reflection of $D'$ over the line $\overline{T_1T_3}$ lies on the line $\overline{H_1H_3}$.

Let $P$ be the reflection of $D'$ over $\overline{T_1T_3}$, $X_1 = \overline{H_2H_3} \cap \overline{T_2T_3}$ (and cyclic permutations), and introduce $Y = \overline{T_1T_3} \cap \overline{D'T_2}$. Now by the [url=https://artofproblemsolving.com/community/c618937h1717975]Right Angles on Incircle Chord Lemma[/url], $\angle AYC = 90^{\circ}$. This implies that both $AH_3H_1YC$ and $T_2IT_1YC$ are cyclic.

The crux of the problem lies in the following:

[b][color=#f00]Claim.[/b][/color] $T_1$ and $P$ are symmetric around $H_1Y$.

[i]Proof.[/i] Notice that $YT_1 = YD' = YP$ by symmetry and definition of $P$. Therefore, it suffices to show that $YH_1$ bisects the angle $PYT_1$. Notice that
\begin{align*}
\angle H_1YT_1 &= \angle H_1YA - \angle T_1YA \\
&= \angle H_1CA - \angle T_1CI \\
&=\frac C2,
\end{align*}
while
\begin{align*}
\angle PYT_1 &= \angle T_1YD' \\
&= 180^{\circ} - 2\angle T_1T_3T_2 \\
&= 180^{\circ} - (180^{\circ} - C) \\
&= C,
\end{align*}
proving the claim. $\blacksquare$

This means that $H_1T_1P$ is isosceles with $H_1P = T_1P$. Observe that
\begin{align*}
\angle H_1T_1P &= 360^{\circ} - 2\angle D'T_1Y - \angle H_1T_1D' \\
&=360^{\circ} - (180^{\circ} - C) - \left(180^{\circ} - \frac 12 \widehat{DT_1}\right) \\
&= C + \frac 12(360^{\circ} - 2\widehat{T_1T_3} - \widehat{T_2T_3}) \\
&= C+\frac 12(360^{\circ} - 2(180^{\circ} - B) - (180^{\circ} - A)) \\
&= C + \frac 12(2B+A-180^{\circ}) \\
&= \frac B2 + \frac C2 \\
&= 90^{\circ} - \frac A2,
\end{align*}
implying that $\angle T_1H_1P = A$. But $\angle H_3H_1T_1 = 180^{\circ} - A$ by cyclic $AH_3H_1C$ implies $\angle H_3H_1T_1 + \angle PH_1T_1 = 180^{\circ}$, so $P, H_1, H_3$ are collinear, as required. $\square$

[b][color=#9a00ff]Remark.[/color][/b] By no means is this specific to this problem only, but I discovered that a lot of motivation for constructing points comes from completing the configuration -- in this problem, $Y$ completes the incircle configuration. Now, we were able to use the "dual" properties of $Y$ (that it lies on the two lines from which it is defined, and the two right angles) to help solve the problem.

[img]https://cdn.discordapp.com/attachments/872490629714808853/900905179275198464/unknown.png[/img]

Beautiful problem :love:

Let $H_1'$ and $H_2'$ be the reflections of $H_1$ and $H_2$ over $\overline{T_1T_2}$, respectively, and let $T_1'$ be the reflection of $T_1$ over the perpendicular bisector of $\overline{T_2T_3}$. It's sufficient to prove that $H_1'$, $H_2'$, and $T_1'$ are collinear.

[b][color=#00f]Claim 1:[/color][/b] $H_1'$ lies on the angle bisector of $\angle BAC$.
[i]Proof:[/i] Let $X$ be the concurrency point of $\overline{T_1T_2}$, $\overline{T_1'T_3}$, and the angle bisector of $\angle BAC$. By the Iran Lemma, $\angle AXB=90^\circ$, so $AH_2XH_1B$ is cyclic. Thus, we have \[\measuredangle AXT_2=\measuredangle AT_2X+\measuredangle XAT_2=\measuredangle XT_1H_1+\measuredangle BAX=\measuredangle XT_1H_1+\measuredangle T_1H_1X=\measuredangle T_1XH_1H_1'XT_1,\] so $A$, $X$, and $H_1'$ are collinear.

[b][color=#00f]Claim 2:[/color][/b] $\overline{AB} \parallel \overline{H_1'T_1'}$
[i]Proof:[/i] We have \[\measuredangle XH_1'T_1'=\measuredangle T_1H_1'X=\measuredangle XH_1T_1=\measuredangle XAB,\] so $\overline{AB} \parallel \overline{H_1'T_1'}$.

[b][color=#00f]Claim 3:[/color][/b] $\overline{AB} \parallel \overline{H_1'H_2'}$
[i]Proof:[/i] Notice that $\overline{H_1H_2}$ and $\overline{H_1'H_2'}$ are antiparallel with respect to $\angle ACB$. Since $AH_2H_1B$ is cyclic, we know that $\overline{AB}$ and $\overline{H_1H_2}$ are antiparallel with respect to $\angle ACB$, so our claim is proven.

Combining claims 2 and 3 gives the desired result.

Let $G_i$ be the reflection of $H_i$ over $\overline{T_iT_{i+1}}$ for $i=1,2,3$ where $T_4=T_1.$ Similarly, let $J_i$ be the reflection of $H_i$ over $\overline{T_{i-1}T_i}$ for $i=1,2,3$ where $T_0=T_3.$ We use complex numbers with unit circle $(T_1T_2T_3).$ Let $d=t_1,e=t_2,$ and $f=t_3.$ Notice that $H_1$ is the foot from $A=\frac{2}{\overline{e}+\overline{f}}=\frac{2ef}{e+f}$ to $\overline{T_1T_1}$ so $$h_1=\frac{1}{2}(2d+a-d^2\overline{a})=\frac{de+ef+fd-d^2}{e+f}.$$ Also, $G_1$ is the reflection of $H_1$ over $\overline{DE}$ so \begin{align*}g_1&=\frac{(d-e)\overline{h_1}+\overline{d}e-d\overline{e}}{\overline{d}-\overline{e}}\\&=d+e-ed\overline{h_1}\\&=d+e-ed\left(\frac{\frac{1}{de}+\frac{1}{ef}+\frac{1}{fd}-\frac{1}{d^2}}{\frac{1}{e}+\frac{1}{f}}\right)\\&=\frac{(d+e)(e+f)d-de(d+e+f)-ef}{(e+f)d}\\&=\frac{(d^2+e^2)f}{d(e+f)}.\end{align*} Similarly, $j_2=\frac{(d^2+e^2)f}{e(f+d)}.$ We claim that $G_1,J_2,$ and $P=\frac{ef}{d}$ are collinear. Indeed, $$p=\lambda g_1+(1-\lambda)j_2$$ where $\lambda=\frac{(e+f)(d^3-e^2f)}{(d-e)(d^2+e^2)f}$ and $\lambda\in\mathbb{R}$ as $$\overline{\lambda}=\frac{\left(\frac{1}{e}+\frac{1}{f}\right)\left(\frac{1}{d^3}-\frac{1}{e^2f}\right)}{\left(\frac{1}{d}-\frac{1}{e}\right)\left(\frac{1}{d^2}+\frac{1}{e^2}\right)\frac{1}{f}}=\frac{\frac{e+f}{ef}\cdot\frac{e^2f-d^3}{d^3e^2f}}{\frac{e-d}{de}\cdot\frac{d^2+e^2}{d^2e^2f}}=\lambda.$$ Similarly, $P,G_2,$ and $J_3$ are collinear. Hence, $P=\overline{G_1J_2}\cap\overline{G_2J_3}$ and note that it lies on the unit circle. Similarly, the other intersections of the reflections of $\overline{H_1H_2},\overline{H_2H_3},$ and $\overline{H_3H_1}$ lie on $(T_1T_2T_3).$ $\square$

Let $I$ be the incenter. Throughout the solution we denote by $\mathcal {S}_i$ reflection over line $T_{i-1}T_{i+1}$ (indicates are taken modulo $3$). Our goal is to prove that $\mathcal {S}_{3} (H_1H_2),\mathcal {S}_{2} (H_1H_3)$ concur on incircle; then by analogous assertions we are done. We claim, that the concurrence point is reflection $R$ of $T_1$ over $AI.$

[b]Claim.[/b] $\mathcal{S}_3 (H_1)\in AI.$
[b]Proof.[/b] Let $AI\cap T_1T_2=P, BI\cap T_1T_2=Q.$ By the Iran lemma $A,B,H_1,H_2,P,Q$ are concyclic and $$\measuredangle BP\mathcal{S}_3 (H_1)=\measuredangle BPH_1+ 2\measuredangle H_1PT_1 =\measuredangle BAH_1+ 2\measuredangle H_1BQ=$$ $$=\measuredangle BH_1A=\measuredangle BPA\implies \mathcal{S}_3 (H_1)\in AI.$$

Now observe that $\mathcal{S}_3 (T_1H_1)\parallel AC,$ so by reflection over $AI$ we deduce $R\mathcal{S}_3 (H_1)\parallel AB.$ But by reflection over $T_1T_2$ we deduce $\mathcal{S}_3 (H_1H_2)\parallel AB\implies R\in \mathcal{S}_3 (H_1H_2).$ Analogously $R\in \mathcal{S}_3 (H_1H_3),$ so done.

[hide=Diagram]
[img width=50]https://media.discordapp.net/attachments/986049049834713139/1058198020086321264/Screen_Shot_2022-12-29_at_20.40.55.png?width=984&height=1027[/img]
[/hide]

Let $E$ be the point on the incircle such that $ET_1\parallel T_2T_3$ and $F$ be the point on the incircle such that $FT_3\parallel T_1T_2.$ We claim that $EF$ is the reflection of $H_1H_3$ across $T_1T_3.$ Note that if we prove this, the analogous is also true, which solves the problem. Let the incircle be the $|z|=1$ on the complex plane, and let $T_3,T_1$ be $x=\text{cis}(\alpha),\overline{x}=\text{cis}(-\alpha)$ respectively for some complex number $x$. Clearly, $IB$ becomes simply the real axis. Let $T_2$ be $y=\text{cis}(\beta)$.
\\$~$\\
First, let us determine $E$ and $F$. By the parallel condition, arcs $T_2E$ and $T_2F$ must both be congruent to arc $T_1T_3$. Thus, $\{E,F\}=\text{cis}(\beta\pm 2\alpha).$ We'll compute the intersection of $EF$ with the real axis. If $(a,b)$ and $(c,d)$ are points, then the real-intercept is \[\frac{bc-ad}{b-d}=\frac{\sin(\beta+2\alpha)\cos(\beta-2\alpha)-\cos(\beta+2\alpha)\sin(\beta-2\alpha)}{\sin(\beta+2\alpha)-\sin(\beta-2\alpha)}=\frac{\sin(4\alpha)}{2\sin(2\alpha)\cos(\beta)}=\frac{\cos(2\alpha)}{\cos(\beta)}\]
Now, the real-intercept of $T_1T_3$ is just $\cos(\alpha).$ We need to calculate the real-intercept of $H_1H_3$, and to do so we shall first calculate $IB$, which is equal to $\tfrac{1}{\cos(\alpha)}.$ Next, let $H_1H_3$ and $IB$ intersect at $X$, then
\begin{align*}BX &= BH_1\cdot \frac{\sin(\angle A)}{\sin(\angle C+\angle B/2)}\\
&= BA\cdot \cos(2\alpha)\cdot \frac{\sin(\beta-\alpha)}{\cos(\beta)}\\
&= \left(\tan(\alpha)+\tan\left(\frac{\beta-\alpha}{2}\right)\right)\cdot \frac{\cos(2\alpha)}{\cos{\beta}}\cdot \sin(\beta-\alpha)\\
&= \sin(\beta-\alpha)\left(\frac{\sin(\alpha)}{\cos(\alpha)}+\frac{\sin(\beta-\alpha)}{1+\cos(\beta-\alpha)}\right)\cdot \frac{\cos(2\alpha)}{\cos{\beta}} \\
&= \frac{\cos(2\alpha)}{\cos{\beta}}\cdot \sin(\beta-\alpha)\cdot \frac{\sin(\alpha)+\sin(\beta)}{\cos(\alpha)(1+\cos(\beta-\alpha))} \\
&= \cos(2\alpha)\left(\frac{1}{\cos(\beta)}-\frac{1}{\cos(\alpha)}\right)
\end{align*}
So we have \[IX=\frac{1}{\cos(\alpha)}-\frac{\cos(2\alpha)}{\cos(\beta)}+\frac{\cos(2\alpha)}{\cos(\alpha)}=2\cos(\alpha)-\frac{\cos(2\alpha)}{\cos(\beta)}.\] Now, let's put it all together. Let $H_1H_3$, $T_1T_3$ and $H_1H_3$ intersect $IB$ at $X,Y,Z$ respectively. We have $X=2Y-Z$ which means that $X$ and $Z$ are reflections across $Y.$ Since $T_1T_3$ is perpendicular to $IB$, it just remains to show that $\angle FZE=\angle H_3XI.$ That is just simple angle chasing. We are done.

This was a failure even after 5 months of try, finally secured it with a hint.

[quote=SL2000 G8]Let $ AH_1, BH_2, CH_3$ be the altitudes of an acute angled triangle $ ABC$. Its incircle touches the sides $ BC, AC$ and $ AB$ at $ T_1, T_2$ and $ T_3$ respectively. Consider the symmetric images  of  the  lines $ H_1H_2, H_2H_3$ and $ H_3H_1$ with  respect  to  the  lines $ T_1T_2, T_2T_3$ and $ T_3T_1$.  Prove  that  these  images  form a  triangle  whose  vertices  lie  on  the incircle of $ ABC$.[/quote]

Define $D,E,F \in \omega$ where $\omega$ is the incircle such that $T_1D \parallel T_2T_3$ and other variations hold.

[b]Claim : [/b] $DE || AB$ and other cyclical variations hold.
[b]Proof [/b] Kinda note that $D,E$ are reflections of $T_1,T_2$ over perpendicular bisectors of $T_2T_3$ and $T_1T3$ and other trivial things which hold.
So,
$$\angle DET_3=\angle DT_2T_3=\angle T_1T_2T_3=\angle ET_1T_3=\angle EDT_3 \cdots (1)$$
Now we have ,
$$\angle BT_3D=\angle T_3T_1D=\angle T_3ED \stackrel{\text{By (1)}}=\angle T_3DE$$
Which proves our claim $\blacksquare$

[b]Claim : [/b] The reflections of $H_1$ over $T_1T_2$ and $T_1T_3$ lie on $AI$ where $I$ is the incentre and other variations hold too .
[b] Proof : [/b] Say $G$ be the point such that $BG \perp AI$ by Iran lemma $G \in T_1T_2$.Now $GT_2=GT_3$  and $BH_1GA$ cyclic.A little bit of angle chasing with this facts give that if $X$ is a point on $AG$ extended,$GT_1(T_1T_2)$ bisects $\angle H_1GX$.This gives that the reflection of $H_1$ over $T_1T_2$ lies on $AI$ and similarly the other reflection lies on $AI$ as well and hence the claim is proved $\blacksquare$

[b] Claim :[/b]Let $H_1',H_2'$ be reflections of $H_1,H_2$ over $T_1T_2$, we have $H_1',H_2' \in DE$
[b] Proof :[/b] Note that $T_3D \cap T_1T_2=G$ since $AG$ is the perpendicular biseector of $T_2T_3$.
Now
 $$\angle H_1'DT_3=\angle H_1'T_1G=\angle H_1T_1G=\angle BT_1T_2=\angle AT_2T_1=\angle AT_3D $$
The above observation implies $H_1'D \parallel AB \implies H_1' \in DE$.Similarly $H_2'E \parallel AB \implies H_2' \in DE $      $\blacksquare$

So the reflection of $H_1H_2$ over $T_1T_2$ is nothing but the line $DE$,similarly $EF$ and $FD$ are the other reflections.Since all the points $D,E,F$ lie on the incircle of $\triangle ABC$ we're done $\blacksquare$

Notice how reflecting $H_1H_2$ over $T_1T_2$ gives a line with the same direction as if we reclected it over $CI.$ But this just gives the direction of $AB.$ So this tells us that the desired triangle will be similar to $ABC.$ This motivates us to deffine $W_B$ and $W_C$ as the midpoint of arcs $AC$ and $BC$ containing $B$ and $A$ respectively, and $X_1=H_1H_2 \cap T_1T_2,$ $Y_1=AB\cap W_AW_B.$ Now as $W_BW_C \parallel EF$ then by homothety the problem is equivalent to showing that $\frac{Y_1W_A}{Y_1W_B}=\frac{X_1T_1}{X_1T_2},$ which follows easily (say by Menelaus on $\triangle AT_3T_2,$ and by the similarity $\triangle Y_1AW_A \sim \triangle Y_1W_BB$).

Let $\omega$ be the unit circle. Let $W$ be on $\omega$ such that $WT_1 \parallel T_2T_3$. Define $X$ and $Y$ similarly. The following claim finishes since $W$ is on $\omega$.
[b]Claim:[/b] $W$ is where the reflections of $\overline{H_1H_2}$ and $H_3H_1$ intersect. Similar goes for $X$ and $Y$.
[b]Proof:[/b] First note that $w = \frac{t_2t_3}{t_1}$ and $a = \frac{2t_2t_3}{t_2+t_3}$ and similar for $b$ and $c$. Therefore
    \[h_1 = \frac{1}{2}\left(t_1+t_1+\frac{2t_2t_3}{t_2+t_3}-\frac{2t_1^2}{t_2+t_3}\right) = \frac{t_1t_2+t_1t_3 + t_2t_3-t_1^2}{t_2+t_3}\]
    Next, reflecting $H_1$ over $T_1T_2$ yields
    \[h_1' = t_1+t_2-t_1t_2 \overline{h_1} = t_1+t_2 - \frac{t_1^2t_2+t_2^2t_1+t_1t_2t_3 - t_2^2t_3}{t_1(t_2+t_3)} = \frac{t_3(t_1^2+t_2^2)}{t_1(t_2+t_3)}\]
    Similar computation yields
    \[h_2' = \frac{t_3(t_1^2+t_2^2)}{t_2(t_1+t_3)}\]
    Now it suffices to show that $W,H_1',H_2'$ are collinear. This is equivalent to the following being real:
    \[\frac{w-h_1'}{w-h_2'} = \frac{\frac{t_3(t_2t_3-t_1^2)}{t_1(t_2+t_3)}}{\frac{t_3(t_2^2t_3-t_1^3)}{t_1t_2(t_1+t_3)}} = \frac{(t_2t_3-t_1^2)(t_2)(t_1+t_3)}{(t_2+t_3)(t_2^2t_3-t_1^3)}\]
    which is clearly real upon taking the conjugate, so we're done.


[b]Randy[/b]: "Hi Rachel, that's an interesting quadratic equation you have written down.  What are its roots?''
[b]Rachel[/b]: "The roots are two positive integers.  One of the roots is my age, and the other root is the age of my younger brother, Jimmy.''
[b]Randy[/b]: "That is very neat!  Let me see if I can figure out how old you and Jimmy are.  That shouldn't be too difficult since all of your coefficients are integers.  By the way, I notice that the sum of the three coefficients is a prime number.''
[b]Rachel[/b]: "Interesting.  Now figure out how old I am.''
[b]Randy[/b]: "Instead, I will guess your age and substitute it for x in your quadratic equation... darn, that gives me -55, and not 0.''
[b]Rachel[/b]: "Oh, leave me alone!''

Prove that Jimmy is two years old, and determine Rachel's age.

Very nice question!!! :)

well consider the equitation being ax²+bx+c=0
with roots k1 and k2 (k1>k2)

as it are ages, -b/a = k1+k2 thus b is negative
c/a = k1.k2 is positive
(rewrite so that a is positive)

b is negative but is in an integer??  :?
so we'll have to write it as ax²-bx+c=0 (which is important for the summations)

as the solutions are integers, we can write the equitation in a way that a=1
(b=a(k1+k2) and c=a(k1.k2) so they're all divisionable by a) -- (in fact a was 1 from the beginning, as the sum is prime :D)

so x²-bx+c=0
b²-4c here is (k1-k2)²

1+b+c is odd, due to parity stuff k1 and k2 are both even.

so we can rewrite the equitation again, as x²-2dx+4e=0
then the roots: d \pm \sqrt(d²-4e) so the only rule is that d²-4e must have integer roots.

combining with the last unused given: wrong guess gives -55
we get that  \sqrt(d²-16e-220) has integer roots

==> so the only equitation fulfilling all requirements is x²-20x+36=0
so Jimmy is 2 yrs old, and Rachel is 18. And not 13 like Randy guessed.


$ x,y,z \ge 0$
then:
$ \sum_{cyc}x(2x \minus{} y \minus{} z)(x \minus{} y)(x \minus{} z) \ge 0$ :?:

[quote="zhaobin"]$x,y,z \ge 0$
then:
$\sum_{cyc}x(2x-y-z)(x-y)(x-z) \ge 0$ :?:[/quote]

Fails for x = 2, y = 2, z = 1.

  darij


Let x, y, z >0 and xyz=1
prove
$x+y+z\le xy+yz+xz$
 :maybe:


topology
function
real analysis
real analysis unsolved

Consider $X$ the space of sequences in $[0,1]$ with the distance given by $d(x,y)=\sum {\frac{|x_n-y_n|}{2^n}}$. Prove that any compact metric space is homeomorphic to a subspace of $X$. Can we choose this subspace to be closed?

One thing I can't understand: what's the point of the last question? :? Since $X$ is a Hausdorff topological space, every compact subspace [b]must[/b] be closed, so every embedding of a compact metric space into $X$ will certainly be a closed subspace of $X$. This makes the question "[b]can[/b] we choose it to be closed?" look rather silly :?. Am I missing something here?

Edit:

Let $M$ be a compact metric space. We remetrize it, replacing the original metric $\rho$ in $M$ with $\frac{\rho}{1+\rho}$ to make it bounded above by $1$. We will also denote this new distance by $\rho$, to keep things simple. Let $(m_n)_n\subset M$ be a countable dense subset of $M$. Now map each $m\in M$ onto the sequence of distances $(\rho(m,m_n))_n$, which is obviously an element of $X$.

This map (which we will denote by $f$) is injective, because for every $m\in M$ there is a subsequence $(m_{k_n})_n\to m$, and if we have $(\rho(p,m_{k_n}))_n\to 0$ and $(\rho(q,m_{k_n}))_n\to 0$ for some $p,q\in M$, then $p$ and $q$ are limits of $(m_{k_n})_n$ in $M$, meaning that they must coincide. $f$ is continuous because if $(p_k)_k\to p$, then $(\rho(p_k,m_n))_k\to \rho(p,m_n)$ uniformly with respect to $n$, and we get $\left(\sum_n\frac{|\rho(p_k,m_n)-\rho(p,m_n)|}{2^n}\right)_k\to 0$.

We’ve shown that $f$ is a continuous bijection from a compact topological space $M$ onto a Hausdorff topological space $f(M)$. It is a very well-known result that such functions are actually homeomorphisms, and we’re done.

Interesting, the only proof of the "silly" question I asked has 9 pages and appear in a quite modern and advanced topology book...  ;)

Could you please explain then what it is that I'm overlooking? :? What I am saying is this: a compact subset of a Hausdorff topological space is closed, so the question "can we take the subspace to be closed?" seems pointless and awkward.


[hide="Answer"]
Since the roots are $x_1$ and $x_2$, we can rewrite the equation as
$(x - x_1)(x - x_2)$
Multiplying this gives
$x^2 - x_2x - x_1x + x_1x_2$
So $m = x_1x_2$ and also $-x_2x - x_1x = x \Rightarrow -x_2 - x_1 = 1$.
We can then use substitution to solve for $x_2$:
$(-1 - x_2)^2 + x_2^2 = 5$
$1+2x_2 + x_2^2 + x_2^2 = 5$
$x_2^2 + x_2 - 2 = 0$
$(x_2 + 2)(x_2 - 1) = 0$
$x_2 = -2, 1$
$x_1$ is then
$-1 - -2 = 1$ or $-1 - 1 = -2$
So $m = (-2)(1)$ or $m = (1)(-2)$ so the answer is $A$[/hide]

Silverfalcon Note: If you don't mind, please hide your answer so other people can't look at it.

Hm, I just used quadratic formula and squared the two solutions.  I got the same answer.

$x_1^2=-(x_1+m), x_2^2=-(x_2+m)$, so $x_1^2+x_2^2=5\Longleftrightarrow -(x_1+x_2)-2m=5\Longleftrightarrow 1-2m=5\Longleftrightarrow m=-2$


Find all functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying: $ x^2 f(x)\plus{}f(1\minus{}x)\equal{}2x\minus{}x^4$ for all $ x \in \mathbb{R}$.

[hide="Warning: Long solution dead ahead"]Plug in $ 1\minus{}x$ for $ x$.

$ (1\minus{}x)^2f(1\minus{}x) \plus{} f(x) \equal{} 2(1\minus{}x) \minus{} (1\minus{}x)^4$

Now divide by $ (1\minus{}x)^2$ so we can subtract and cancel the $ f(1\minus{}x)$ from the original equation. We get

$ f(1\minus{}x) \plus{} \frac{f(x)}{(1\minus{}x)^2} \equal{} \frac{2}{1\minus{}x} \minus{} (1\minus{}x)^2$

Then subtracting it from the original equation we get

$ (f(x))(x^2\minus{}\frac{1}{(1\minus{}x)^2}) \equal{} 2x\minus{}x^4 \minus{} \frac{2}{1\minus{}x} \plus{} (1\minus{}x)^2$

$ f(x) \equal{} \frac{2x\minus{}x^4 \minus{} \frac{2}{1\minus{}x} \plus{} (1\minus{}x)^2}{x^2\minus{}\frac{1}{(1\minus{}x)^2}}$

$ f(x) \equal{} \frac{2}{x\plus{}\frac{1}{1\minus{}x}} \plus{} \frac{(1\minus{}x)^2 \minus{} x^4}{x^2 \minus{} \frac{1}{(1\minus{}x)^2}}$

$ f(x) \equal{} \frac{2}{x\plus{}\frac{1}{1\minus{}x}}  \minus{} \frac{(1\minus{}x\minus{}x^2)(1\minus{}x\plus{}x^2)}{\frac{(1\minus{}x\minus{}x^2)(1\plus{}x\minus{}x^2)}{(1\minus{}x)^2}}$

$ f(x) \equal{} \frac{2}{x\plus{}\frac{1}{1\minus{}x}}  \minus{} \frac{1\minus{}x\plus{}x^2}{\frac{(1\plus{}x\minus{}x^2)}{(1\minus{}x)^2}}$

$ f(x) \equal{} \frac{2}{x\plus{}\frac{1}{1\minus{}x}}  \minus{} \frac{(1\minus{}x\plus{}x^2)(1\minus{}x)^2}{(1\plus{}x\minus{}x^2)}$

I probably made some mistakes.. somewhere and it's kind of a long function so I don't feel like plugging it back in lol.[/hide]

[hide]Let $ P(x)$ be the property $ x^2f(x) \plus{} f(1 \minus{} x) \equal{} 2x \minus{} x^4$

$ P(1 \minus{} x)$ yields:
$ (1 \minus{} x)^2f(1 \minus{} x) \plus{} f(x) \equal{} 2(1 \minus{} x) \minus{} (1 \minus{} x)^4$

Multiply $ P(x)$ by $ (1 \minus{} x)^2$:
$ (1 \minus{} x)^2x^2f(x) \plus{} (1 \minus{} x)^2f(1 \minus{} x) \equal{} (2x \minus{} x^4)(1 \minus{} x^2)$

Now subtract $ P(1 \minus{} x)$ from $ P(x)$ times $ (1 \minus{} x)^2$:
$ \left[(1 \minus{} x)^2x^2 \minus{} 1\right]f(x) \equal{} \left[(2x \minus{} x^4)(1 \minus{} x)^2\right] \minus{} \left[2(1 \minus{} x) \minus{} (1 \minus{} x)^4\right]$

$ ((x^2 \minus{} x)^2 \minus{} 1)f(x) \equal{} (1 \minus{} x)\left[x^5 \minus{} x^4 \minus{} x^3 \plus{} x^2 \minus{} x \minus{} 1\right]$

$ (x^2 \minus{} x \minus{} 1)(x^2 \minus{} x \plus{} 1)f(x) \equal{} (1 \minus{} x)(1 \plus{} x)(x^2 \minus{} x \minus{} 1)(x^2 \minus{} x \plus{} 1)$

Therefore: $ f(x) \equal{} (1 \minus{} x)(1 \plus{} x) \equal{} 1 \minus{} x^2$ (which does satisfy the original equation)
for all $ x \in \mathbb{R}$ such that $ (x^2 \minus{} x \minus{} 1)(x^2 \minus{} x \plus{} 1) \neq 0 \Rightarrow x \not\in \left\{\dfrac{1 \minus{} \sqrt {5}}{2} , \dfrac{1 \plus{} \sqrt {5}}{2} \right\} \equal{} \left\{ 1 \minus{} \phi , \phi \right\}$.

Let $ f(\phi) \equal{} a$ where $ a \in \mathbb{R}$ is an arbitrary constant.
Then $ f(1 \minus{} \phi) \equal{} [2\phi \minus{} \phi^4] \minus{} [\phi^2f(\phi)] \equal{} \minus{} (\phi \plus{} 2 \plus{} a\phi^2)$

Therefore the solutions are:
$ f(x) \equal{} 1 \minus{} x^2$ for all $ x \not\in \left\{ 1 \minus{} \phi , \phi \right\}$, $ f(\phi) \equal{} a$, and $ f(1 \minus{} \phi) \equal{} \minus{} (\phi \plus{} 2 \plus{} a\phi^2)$. [/hide]


For those of you who have taken EPGY courses (like Multivariable Differential Calc, etc.), how long did it take you to complete the course?

I have always found that EPGY courses can be completed in much less time than the limit they give you.

I remember in grade 4 i completed gade 5,6,7, and 8 courses in about 6 months. I can't recall many of the details though.

I took EPGY every morning for half an hour before school when I was in first grade.  I remember it being very helpful and it was the best thing I had until I found AoPS.  At that semi-normal pace, I finished the 5th and 6th grade courses in a semester.  One great thing about it is that you can really choose the pace you want.

@ illusiat. EPGY stands for Education Program for Gifted Youth.

I've taken a wide variety of courses from EPGY. I've finished 5th-Calculus I in math, Physics I, English I, and all of the available computer science courses. In my opinion, I prefer the math, Physics and computer science, as they are self-paced and very informative. I did not really enjoy the English class, as we had to go to a virtual classroom once a week, which was very demanding to me. I was usually able to complete the self-paced courses in 2-3 months.

Unfortunately, EPGY does not have any more science courses, so I am taking Biology and Chemistry from CTY. Does anyone know how good the TIP programs are?


Let $ ABCD$ be a cyclic quadrilateral. Points $ M, N,P$, and $ Q$ are chosen on the sides $ AB, BC, CD$, and $ DA$, respectively, such that $ \frac{MA}{MB} \equal{} \frac{PD}{PC} \equal{} \frac{AD}{BC}$ and $ \frac{QA}{QD}\equal{} \frac{NB}{NC} \equal{} \frac{AB}{CD}$.
Prove that $ MP$ is perpendicular to $ NQ$.

MP and NQ are the bisectors of the angles between AC and BD.

Best regards,
sunken rock


suppose that [b]n,d[/b] is a natural number and [b]2n 2[/b] divisible by [b]d[/b].prove that
[b]n 2 +d[/b] is not a perfect square.


60m^4 - 120m^3n + 50m^2n^2=

How can one make factors of this? I mean cut down the expression, called factorizing where im from..

 :huh:

You mean $ 60m^{4}\minus{}120m^{3}n\plus{}50m^{2}n^{2}$, right?

Well...from what I can tell, you can factor out $ 10m^{2}$ from each term, giving you: $ 10m^{2}(6m^{2}\minus{}12mn\plus{}5n^{2})$ I think you could take this one step further, but I have to go...sorry.

Yes, but its very similar to the:

a^2 - 2ab + b^2 = (a-b)^2

but i dont find a root that fits 60 or 6...

I think that's as far as you can go with factoring...unless you want to go to irrationals.

[hide="WRONG Irrationals"]
EDITTED Multiplied by mistake
$ 10m^{2}\left( m\minus{}\left(\frac{12n\plus{}n\sqrt{24}}{12}\right)\right)\left( m\minus{}\left(\frac{12 n\minus{}n\sqrt{24}}{12}\right)\right)$
$ 10m^{2}\left( m\minus{}n\minus{}\frac{n\sqrt{6}}{6}\right)\left( m\minus{}n\plus{}\frac{n\sqrt{6}}{6}\right)$.
[/hide]

Out of curiosity, can you show me how you worked that out? It doesn't seem like the quadratic formula would work in this situation.

Used the quadratic and just added the $ n$ part of the expression.
Do not know if it is correct though after looking at it again.

Oh duh...just treat $ n$ as if it were a number...I hate how I miss such simple concepts so easily.


There are 3 yellow, 4 blue, and 5 green marbles in a bag. You randomly draw marbles one at a time from the bag. What is the probability that you will draw all three yellow marbles before you draw a green one?

Can you generalize for x yellow, y blue, and z green marbles?

[hide=""for the question""]

$\frac{3}{12} \cdot \frac{2}{11} \cdot \frac{1}{10} = \frac{1}{220}$?
[/hide]

[hide="generalization"]
If we have x yellow, y blue, z green, then the probability of drawing all x yellow before puling out a green one is:

$\frac{x}{x+y+z} \cdot \frac{x-1}{x+y+z-1} \cdot.... = \frac{x!}{(x+y+z)!}$
[/hide]

You keep on drawing even if you've already drawn all 3 yellow marbles.

reddevil, you cannot put colons in the title of a Hide tag.  I fixed it for you.  Just thought you might want to know for future reference.


function
induction
algebra unsolved
algebra

Prove that there is no function f such that f:R+->R+, f(x+y)>f(x)(1+yf(x)) for all positive real numbers x,y

Suppose $ f(a_{0})\equal{}b_{0}$, and let $ a_{n\plus{}1}\equal{}a_{n}\plus{}\frac{1}{b_{n}}$ and $ b_{n}\equal{}f(a_{n})$. Then it is easy to see by induction that $ b_{n\plus{}1}>2b_{n}$, as $ b_{n\plus{}1}\equal{}f(a_{n}\plus{}\frac{1}{b_{n}})>f(a_{n})(1\plus{}\frac{f(a_{n})}{b_{n}})\equal{}2b_{n}$. But then notice $ a_{n\plus{}1}\equal{}a_{0}\plus{}\frac{1}{b_{0}}\plus{}\frac{1}{b_{1}}\plus{}...\plus{}\frac{1}{b_{n}}<a_{0}\plus{}\frac{1}{b_{0}}\plus{}\frac{1}{2b_{0}}\plus{}\frac{1}{4b_{0}}\plus{}\frac{1}{8b_{0}}\plus{}...\plus{}\frac{1}{2^{n}b_{0}}\equal{}a_{0}\plus{}\frac{2}{b_{0}}$, so $ a_{n}$ converges and obviously $ b_{n}$ does not. But since $ f(n)$ is clearly monotonic increasing, and it approaches infinity as $ n$ approaches some constant $ c$, certainly $ f(c)$ is undefined.


function
complex analysis
complex analysis unsolved

Suppose that we have an open bounded set $ D$ and $ C\setminus { \overline {D} }$ is  connected. Can we construct a sequence of conformal mappings $ \phi_n$ such that $ \phi_n : A_n\to B_n$ where $ B_n\subset D\subset A_n$ and $ \phi_n\to z$ uniformly on $ D$ when $ n\to\infty$? (one can first suppose that $ D$ is connected)
 Maybe this is not  very rigirous formulation of what I want to ask,sorry if it's stupid question :)

The disk with a cut is a clear counterexample if you want $ B_n\supset \overline D$. As written, the question hardly makes any sense at all because nothing prevents us from taking $ A_n\equal{}B_n\equal{}D$, $ \varphi_n(z)\equal{}z$. So, at this point, the question is just poorly stated :(.

I knew that the problem would be poorly posted :blush:
Well, actually I want that $ B_n\subset D\subset A_n$  (strict embending) where $ A_n$ and $ B_n$ are conformaly isomorphic.
I also have the function $ f$  analytic on  $ D$ and continuos on  $ \overline D$, so  for any $ D$ with ''nice'' cut everything is fine (I mean the cut can be dropped ).
Anyway, maybe you know what conditions(on $ D$)  I should add for the problem to be correct?
Thanks.


What did everyone think of Lord of the Rings the third movie? I thought the action was great - almost too unrealistic though. It was hanging there on the boundary, so I still enjoyed it. I thought the movie was rather prolonged beyond necessity, and the ending was slow. The 3 and a half hours was long, like I said before. However, putting the whole series together, the LOTR was really an enjoyable trilogy that I will probably watch again. Personally I liked the second one the best, but the green ghosts in the third movie did create a special effect. Which one of the three did you like the best. Discuss your reasons for choosing so.

ok, i think ROTK was the best because of the battle scenes... that was just way too cool... seriously, like 100,000 people/orcs/other things i guess fighting was beyond imagination... and the birds-eye view of sauron's army was crazy, cept i could make a debate on how minus tirith or sauron's army could've easily defeated the other if they actually planned ahead, like more fire, better gates, etc. and i dont get why the riders of rohan were able to run over all those orcs... they shouldve shot arrows at the horses and just killed them all b4 the riders could even reach the orcs... anyway, ROTK was the best cept for the extended and useless ending.

I agree with the ending. I just voted RoTK since it's the ending and it's like the only one where everyone's happy-happy and all but to tell you the truth [hide]I've seen better movies[/hide]

Why not the books?  Can we vote for the book we liked best instead of the movie I don't watch movies I read.  So ca we vote for our favorite book in the trilogy?

RoTK was great.. but a few point-outs:
1) Why on earth were the dead people bright green?
2) Was it just me, or did frodo get back all his fingers at the end of the movie?
3) Battle scenes were pretty much the same as in TTT, they definitely could've done better

But overall, it was pretty good...
heh... elfstone and evenstar reunion at the end was sweet... =)

yeah there was a really bad scene where you can see his finger bended over. also notice that his main scar on his face changes positions MAJORLY during the last bit in the movie (yeah i saw the movie three times during the first week!)

I personally can't wait to own the entire trilogy on DVD with all the extra footage.  This trilogy needed to be very long in order to include most of the great parts of the books.  I wish I could see all three extended versions at a theater.

I could make all kinds of minor criticisms about the movies, but overall I think an excellent job was done.

[quote="BHorseMath"]I really haven't watched much of the movies but I know the books backwards and forwards.   From what I have seen of the movies though they follow the books pretty good where as most movies don't.[/quote]

There are definitely some major departures from the books.  Tom Bombadil is entirely absent from the movies, but that makes sense when you have to cut time because he's not essential to the main plot.

There are a few characters who are given very different personalities in the movies than the books and that's one of my biggest criticisms.  However, without some changes to please modern audiences, the movies never would have been made, so I can put up with it.

Yes my dad said that ther were those kind of things.  Thats why I don't watch movies.  I like to imagin things and personalities and apperences.  I read the book about 3 years ago because I wanted a good harder book and my dad told me it was a good one he read it in high school and thought that I might understand it.

Heh, I went to the thing in NY where they showed all three back to back.. it was awesome!! =) =) =)

The LoTR series is not to be read just for a "harder book"... you should read it just for enjoyment too... and don't forget the Silmarillian and Unfinished Tales while you're at it, it'd be awesome if they did movies for all the annals of history.. hehe

yeah I have read parts of those books and I didn't just read it for being harder but because I was looking for something more enjoyablre and a book that would take me longer than a couple hour to finish.  As it was the books only took me a few days.

I was so sad they cut out Bombadil -- I think he was my favorite character :)

Wow, you're a fast reader...took [b]me [/b]a couple of [b]months[/b]! :lol:
[quote="lunmu"]Heh, I went to the thing in NY where they showed all three back to back.. it was awesome!! =) =) =)[/quote]
Whoa, you're a die hard fan...I would like die and my eyeballs will rot if I had to watch all of em, isn't that like 7 hours long? That's like watching all three Matrix or Terminator or Godfather at one time...then again that may not be a bad idea :)

I'm not that fast of a reader Tare I just read that series fast The Harry Potter series took me 3 whole days to read all of the back to back 1-5.

Yeah I was upset that they bombadil.  He has a really good character.

[quote="lunmu"]... and don't forget the Silmarillian and Unfinished Tales while you're at it, it'd be awesome if they did movies for all the annals of history.. hehe[/quote]

I personally never finished the Silmarillian.  It's a history and I didn't find it nearly as interesting or entertaining as the LOTR.  In fact, I haven't yet finished it for that reason.  I'm not at all sure it would make a good film.

It might not be as interesting, but it really helped me understand the whole story behind LoTR better...

I love LOTR. The plot was so creative that Terry Brooks praticallly copy the whole main idead into the The Sword of Shmilaron. (If you read it you'll wonder if there is plagarism)

I have read part of that series and while the main idea is somewhat similar in my opinion it is a whole different book.  The characters and plots are pretty different from LOTR.


suppose  that : $ \minus{} 1\le ax^{2} \plus{} bx \plus{} c\le 1$  for all  $ \minus{}1\le x\le 1$  then show that
$ \minus{} 4\le 2ax\ \plus{} b\le 4$ for all  $ \minus{} 1\le x \le 1$
where a,b,c are real.

[quote="tamal banerjee"]suppose  that : $ - 1\le ax^{2} + bx + c\le 1$  for all  $ - 1\le x\le 1$  then show that
$ - 4\le 2ax\ + b\le 4$ for all  $ - 1\le x \le 1$
where a,b,c are real.[/quote]

noitice that $ f^'(X) \in [-4;4]$

[quote="tamal banerjee"]suppose  that : $ \minus{} 1\le ax^{2} \plus{} bx \plus{} c\le 1$  for all  $ \minus{} 1\le x\le 1$  then show that
$ \minus{} 4\le 2ax\ \plus{} b\le 4$ for all  $ \minus{} 1\le x \le 1$
where a,b,c are real.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=51487


geometry
topology
abstract algebra
real analysis
number theory
rudin

But In India we do get cheaper editions(in price ofcourse) of springer   :P

[quote="Soarer"]What is your opinion on the price of math books?

Personally I think it is really too expensive. My family is not rich, and I can't really afford those expensive books (especially Springer ones!). Do you think the publishers should lower the cost of the books so that more people will be able to buy them?[/quote]

Look, here in Argentina 1 dollar=3 pesos (our money is called "pesos"). Then, all foreign books are really expensive. That's why Internet is important for us, because I, personally, would never have good books...... 1º because they're expensive, and 2º because they never bring it here........

I have a quesion. How come international version far cheaper? And also the version of the Fundamental of Physics that translated into Korean costs only 20 bucks while US version is 156 dollars!

[quote="Soarer"]What is your opinion on the price of math books?

Personally I think it is really too expensive. My family is not rich, and I can't really afford those expensive books (especially Springer ones!). Do you think the publishers should lower the cost of the books so that more people will be able to buy them?[/quote]

In North America, Springer texts are usually cheaper than their competition.  They are slightly more expensive than the (excellent) AMS series of texts, and of course they are more expensive than Dover texts.  They are significantly less expensive than Wiley Interscience or McGraw Hill or Academic Press texts.

[quote="gauss202"][quote="Soarer"]What is your opinion on the price of math books?

Personally I think it is really too expensive. My family is not rich, and I can't really afford those expensive books (especially Springer ones!). Do you think the publishers should lower the cost of the books so that more people will be able to buy them?[/quote]

If they lowered the price of the books too much, they may not be able to make a profit - meaning there would be no books at all... for anyone.

When someone opens a business or makes a company they are not doing it out of charity. They are doing it first and foremost to make a profit. Without a profit a company cannot survive. If in the process of making money however, they are providing a useful product or service, then it becomes a win-win for both the provider and (at least some of) the consumers.

Also, the great thing about books is that you can find them in the library! If you live near a University you might be able to find a lot of S.V. books. Or you can hang out for hours in the bookstore like I used to do and read them there.[/quote]

I have entered university already, so yes I can get access to many SV books in library. Yet, I prefer a slow reading so very often it takes me a long tiem to finish a book. But others may want to read it as well and recall my book...

For foreign edition, yes, usually edition published in other places are cheaper. But sometimes the books are still expensive to me.... eg Chinese version of Serge Lang's algebra is around 130 HKD(~17 USD), which is still far more expensive than usual China books.

[quote="fredbel6"]basically I learn from my syllabus at university
they are sometimes very well written and to the point, and our professors just charge copy costs (and sometimes they put the pdf online)

I agree about the libraries, our university doesn't have a huge library, but you would be surprised what they have.  On the other hand, normal libraries aren't usually that interesting, just a few statistics books.

José, is that because of inflation in Argentina (I heard that there are a lot of problems with the currency there)
Well , if you ever go out of Argentina, always go to a book shop, you never know...[/quote]

fredbel, what language are the notes in? I am pretty interested.

For me, I prefer self learning much more to learning through lessons. That's why high book cost is a huge problem for me.

[quote="fredbel6"]You can do self learning with a syllabus too, in fact I sometimes do self learning while in class :D

The notes are in English and or Dutch, sometimes they are online, sometimes I downloaded them from the (nonpublic) uni site, but pdf's are quite small and uploadable

In English I think I can find a course on algebraic geometry, topology, finite geometry (designs and strongly regular graphs, ovals and ovoids, polar spaces) , algebra (module theory)[/quote]

do you mind posting the link here? :)

[quote="shobber"]I think you can come to Mainland China ( Like the 九章出版社 in Beijing ). You can find a lot of cheap books here. For instance, I bought a copy of the 5th eddition of the Number Theory and Its Applications by Rosen with no more than 9 dollars and I think it is already very cheap compared to those books in HongKong.[/quote]

Cool!!!!! Here in Argentina I'd never find those books (at any price). They'll never be here. That's the difference between China and Argentina (however we're better at football, ejejej just joking)

[quote="shobber"]I think you can come to Mainland China ( Like the 九章出版社 in Beijing ). You can find a lot of cheap books here. For instance, I bought a copy of the 5th eddition of the Number Theory and Its Applications by Rosen with no more than 9 dollars and I think it is already very cheap compared to those books in HongKong.[/quote]

What?? That cheap?
The baby rudin I own is mainland edition, and it costs me 35 dollars!

Well, I have never seen a text book more than 35 dollars! Even the original books are cheaper than that! Perhaps last time you came to Beijing, the *Cheap* books were not available. But now I guarantee you that you will be satisfied with the current prise of the books!

[quote="shobber"]Well, I have never seen a text book more than 35 dollars![/quote]

The cheapest book I saw was 20 dollars. The most expensive book I saw cost 75 dollars (225 pesos for me is really expensive)

[quote="fredbel6"]I heard that in Poland (under communist regime) you could buy decent physics book(s!) for like five dollars, and not even did you get some work of Lenin with it too, you HAD to take it with you. :?[/quote]

Well, but I prefer to be in democracy with books of 100 dollars than buying books for only 5 dollars but being communist

[quote="fredbel6"]Well that was my relativity theory professor who told that once.  He probably just had to go to Poland for some conference and decided to check out some books, and then went back to Belgium of course.  I wouldn't mind grabbing those works of Lenin along too ;) , although I am a bit surprised those books were in a language he could comprehend (English? :? )[/quote]

Maybe, because see that also Vinogradov's were translated to spanish

Yes, but I don't know whether they were so English oriented in the East Block
I assume most of them thought of Russian as the language to learn :?

Oh, never mind, my professor told me that story a couple of months ago,but when he went we still had francs here in Belgium (one Euro is about 40 francs).
So that must have been before 1990 (and the Euro started in January 2002).

And the price must have gone up by inflation alone.

So just remember "cheap". :roll:

[quote="Soarer"]What dollars are you guys talking about? :?

In my post (and I guess same as shobber's), that we are referring to RMB, currency of China.[/quote]
Unfortunately, I was talking about the US dollars. But still the books are cheap. My copy of the Application of Elementary Number Theory was about 58 RMB ( 7.3 US dollars ) after discount......

I have some friends who studies abroad. They all love to by their text books in China since they are cheap.

about how much are the cheapest textbooks you can find in the US?


geometry
MATHCOUNTS
number theory
prime numbers

Let us compile a top 10 list for MathLinks. :)

You know you're a MathLinker if . . .

let me throw some that I've just cooked up ...

... if you think 42 is the max. score possible on any test
... if you don't know what multiple-choice means
... if IMO is not the shortcut for  In My Opinion, but for International Mathematics Olympiad
... if you don't think that MathLinks is just a collection of links :P

...if you're up at 3 in the morning solving math problems (instead of doing your major essay due tomorrow.)

If you know that the next Olympic games are going to take place in Mexico.

... (you happen to be a girl and) 4 guys who are in love with you have been trying to make each other withdraw by playing a game involving prime numbers :P :lol: :D :oops: :love:

... Valentin(e)'s Day isn't 14th February, but some other day ;)

... the two people who make you laugh the most happen to be one particular Frenchman and another particular Russian :P

...you absolutely refuse to read a geometry proof that is more than 3 screens long, but half the proofs you see are twice that long :( ;)

And let me copy 2 quotes from the "You know you're an AoPSer if..." thread:

... you know that ;) is a wink (contributed by paladin8)

... you think the number 7 will bring you bad luck.....
(Consider being 7th in TST :D :D :D ) (contributed by wpolly)

[quote="Valiowk"]... (you happen to be a girl and) 4 guys who are in love with you have been trying to make each other withdraw by playing a game involving prime numbers [/quote]

Hehehe, or if you happen to be a girl, then the lastest fashions you're waiting for is the Mathlinks and AoPS t-shirt.

[quote="shobber"]you know you're a Mathlinker if when you see the Mcdonald's, you immediatly came up with Sigma...[/quote]

*looks at Mcdonalds. Thinks about food. Cries out in distress.*

I'll try this one again....

You don't like physics because it's not theoretical enough.

No, I totally agree on the physics one, and I thought I was going to be an engineer.

When you know the perfect numbers.

... If Mathman is your hero (I remember Mathman from one of the Mathcounts problems :))
... If you try to save/destroy(?) the world using mathematical equations
... If you want a calculator for your birthday
... If you get upset if your pajamas are not full of numbers
... If you instinctively count how many different ways to pay your hamburger with what's in your wallet

If you know that napkins are no good for doing math on-they rip too easily.

I agree with the one about thinking "IMO" means "International Mathematical Olympiad" rather than "in my opinion." I saw that happen on another online forum. The guy who "explained" the abbreviation some other user posted was, of course, an AoPS-MathLinker.

[quote="Magnara"]To make it a little more subtle, maybe you should say: "You don't think IMO stands for 'In My Opinion.'"[/quote]

Haha the first time I saw IMO used as "in my opinion" I was so confused...

So that's what it stands for. Every time I saw someone type imo I would get so excited...

But there is no $\phi$ day, because $\phi = 1.68\ldots$

[quote="Farazy"]If your favorite pick up line is  "your eyes look like $\pi$"...
If your name is filletwho... :D
$\pi$ day is an extremely minor holiday compared to $\phi$ day...[/quote]

there's a $\phi$ day? what day is that? (maybe i'm just weird...)

[quote="Farazy"]If your favorite pick up line is  "your eyes look like $\pi$"...
If your name is filletwho... :D
$\pi$ day is an extremely minor holiday compared to $\phi$ day...[/quote]

Finally, I can consider myself a mathlinker.

Oops...yeah, I'm wrong, Lucky707 is right.

Oh well! I am sure its a date on some odd calendar!

But you could also consider $\phi$ day January Sixth.  :D How's that for a celebration! :P

(I just made up  $\phi$ day by the way.)

When someone says Moo, you think 10^-6...

Can you guys please lead your random discussions in another topic or make-up one if there isn't any. This thread has a different topic.  ;)

...If you count the prime numbers to get to sleep instead of counting sheep

...If you do all your math homework in your head before you finally think to write it down ;)

If you think school should be out for  :pi: day

[quote="elvenchamp777"]It should!!! Not because it's very important but i just wanna miss school :D

elvenchamp777[/quote]

I am outraged!!  :pi: day is the most sacred nerd holiday around, and you have no respect for it?

schools don't have respect for anything these days, but anyways...

...if you have dozens of missing assignments in biology and are a month ahead in [whatever math class you have].
...if your trekkie father calls you a geek.
...if you traveled all the way to the museum of science and industry in chicago just to buy the 'pi a la mode' shirt (don't know if they have that anywhere else)
...if you, as a child, drew the platonic solids on your notebooks while everyone else drew stick figures.

if you are going to play "Defend Your Castle" in your TI-89+

my mistake, it was just a TI-84 plus...

...if you get excited about new calculators

...you wish standardized math tests had a "long answer" section.

[quote="hmoonster"]... If you try to save/destroy(?) the world using mathematical equations[/quote]

Now who do you know doesn't try to do that? :D

[quote="2kid*bari2"][quote="hmoonster"]
... If you try to save/destroy(?) the world using mathematical equations[/quote]

Now who do you know doesn't try to do that? :D[/quote]
using mathematical equations? only mathlinkers  can do it.

[quote="dodeca-what now?"]ooo...mathgeek, is that a TI-84 plus silver edition with interchangable face plates?   sooooo cooollll![/quote]
I'm sorry, it's a TI 89 titanium.

darn...

...if you say you have to use the computer to do a report, and then come here as soon as you get on it

[quote="dodeca-what now?"]

...if you say you have to use the computer to do a report, and then come here as soon as you get on it[/quote]

Wow, sooo true.

[quote="randomdragoon"]... if you wait until March 14th, 1:59:26.535897932... PM to hold an infinitesimally short celebration of pi day. The rest of the day you just celebrate "Pi rounded to the nearest hundreth" day.[/quote]

I don't think it would be pi day then... it's more like "Pi Moment." ;)


function
floor function
number theory open
number theory

Is there any combinatorial interpretation of the greatest integer function (floor functio)? If yes, can anyone explain??

If $ x$ is nonnegative, then $ \lfloor x \rfloor$ is the number of positive integers that are less than or equal to $ x$.

And, of course, $ \left\lfloor \frac {n}{p} \right\rfloor$ is the number of positive integer solutions to $ px \le n$.  This is a much more useful principle than it seems.

Thanks can you please give a proof of the last post??

In t0r's post, we should assume that $ p$ is positive and $ n$ is nonnegative.  We can rewrite $ px \le n$ as $ x  \le \frac{n}{p}$.  So t0r's result follows from my post.


algorithm
function
linear algebra
matrix
algebra
polynomial
algebra unsolved

If a discrete function is defined with:

$ x(1,n)\equal{}\sum^M_{i\equal{}1}f(i)\,x(i,n\minus{}1)$

$ x(k\plus{}1,n\plus{}1)\equal{}(1\minus{}f(k))\,x(k,n)$

where $ n > 0$ and $ k \in [1,M]$

Also:

$ x(k,n)\equal{}0 \qquad \forall k\in [1,M],\,n \leq0$

but

$ x(1,0)\equal{}X$

How can I express $ x(n,k)$ in terms of $ n$, $ k$, $ M$ and $ X$? Of course $ n, k, M \in \mathbb{Z}$.

You're asking for way too much.
This is a linear recurrence, which can be given in matrix form $ x_{n\plus{}1}\equal{}Ax_n$, $ A\equal{}\begin{bmatrix}f_1&f_2&f_3&\cdots&f_{M\minus{}1}&f_M\\1\minus{}f_1&0&0&\cdots&0&0 \\0&1\minus{}f_2&0&\cdots&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\cdots&0&0\\0&0&0&\cdots&1\minus{}f_{M\minus{}1}&0\end{bmatrix}$

This is close enough to being a companion matrix that we can get its characteristic polynomial easily: the characteristic polynomial is $ x^n\minus{}f_1x^{n\minus{}1}\plus{}(1\minus{}f_1)f_2x^{n\minus{}2}\plus{}(1\minus{}f_1)(1\minus{}f_2)f_3x^{n\minus{}3}\plus{}\cdots \plus{}(1\minus{}f_1)(1\minus{}f_2)\cdots (1\minus{}f_{M\minus{}1})f_M$.

If you want a general explicit solution, you have to find the roots of an arbitrary polynomial first.
If $ 0< f_i<1$ for $ 1\le i<M$ and $ f_M\equal{}1$, this is a Markov chain, and the dominant eigenvalue $ 1$ has a one-dimensional eigenspace, easily found. The rate at which arbitrary initial conditions decay to this is hard.

I figured it wasn't easy. But what if the function $ f$ would be linear? The thing I'm trying to do actually contains such a function and it equals:

$ f(n)\equal{}\frac{M\plus{}1\minus{}n}{M(M\plus{}1)}$

Would this help somehow in the final solution. Also, permitting the solution to contain product's, could then something constructive be done?

If you know $ f$, you can write down the polynomial explicitly and try to solve it to diagonalize $ A$. It probably won't be possible to do so explicitly for large $ M$.

Last note: unfortunately, I'm not fully acquainted with connecting a characteristic polynomial to it's matrix. I suppose that the solution can be solved in the form:

$ x_{n}\equal{}A^n x_0$

Suppose I have solved the polynomial and have the roots $ x_1, ... , x_M$. How would I proceed to get $ A^n$? BTW, I REALLY appreciate your help.


geometry
analytic geometry
circumcircle
parameterization
geometry unsolved

Let two circles, centered at $ A,B$ intersect at $ P,Q$. A variable chord passing through $ P$ intersects the circle with center $ A$ at $ D$ and intersects the circle with center $ B$ at $ E$. Find the locus of the orthocenter of triangle $ DEQ$.

The locus of orthocenter is locus of points H such that DHE=PDQ+PEQ(angles)

Here is a no very nice solution with coordinates: let P(0,p),Q(0,-p), A(-a,0) , B(b,0) and suppose that the radius of circle with center A is lesser than that of circle of center B.
After computations, the locus of orthocenter is a circle, with center at K(0, (p-p^2-ab)/p) - that is, the center is in the line PQ - and the radius is (p^2-ab)/p.

know lemma: let a triangle ABC, a point P inside and a point D on BC. The circumcircle of BDP meet AB on E and the circumcircle of CDP meet AC on F. Then A stay on the circumcircle of PEF. [proof: some realy easy angle]

The perpendicolar line from E to DQ meet the circle center on B on F and the perpendicolar line from D to EQ meet the circle center on A on C.
Clearly C and F are fixed because $ \angle QEF$ and $ \angle QDC$ are fixed. Let H be the orthocenter of DEQ, so by the lemma H stay on the circumcircle of CEF.

[quote="¬[ƒ(Gabriel)³²¹º]¼"]
 Let H be the orthocenter of DEQ, so by the lemma H stay on the circumcircle of CEF.[/quote]

I'm sorry, but the lemma applied to which triangle? I can't really see..

Also, I think in locus problem you need to prove that all points on the circle really belongs to the asked locus  :)

[quote="discredit"]I'm sorry, but the lemma applied to which triangle? I can't really see..[/quote]
to the triangle DEH

[quote="discredit"]Also, I think in locus problem you need to prove that all points on the circle really belongs to the asked locus  :)[/quote]

yes but in this case is obvious that the line CD meet the circle on all of his points when D is moving on the circle with center A.

[quote="28121941"]Dear ¬[ƒ(Gabriel)³²¹º]¼, in your post number 6, what is the so called point G? Thanks in advance[/quote]
$ G \rightarrow F$

I am attaching in a pdf file the relevant details of my analytic solution. I would be very grateful if anybody points out my possible mistakes. Thanks in advance.

[quote="discredit"][quote="28121941"]In the case in which triangle DEQ is acute, the orthocenter H is internal to triangle. EF is an altitude, and H belong to this line. How is possible H belong to the circumcircle of CEF?[/quote]

I think there's a typo... it must belong to the circumcircle of CQF  :)[/quote]

oh  :blush:  yes, sorry all, it was CQF

[quote="28121941"]Once detected a small mistake in my way of eliminate the parameter, all is OK now. My circle coincides with CFQ.
By the way, if I am not wrong, the lemma used by Gabriel in their solution is the so called theorem of Miquel.[/quote]

Are there other solutions? If so, please post it  :)

As [b][size=100]Gabriel[/size][/b] said, the lines through $ D,\ E$ and perpendicular to $ QE,\ QD$ respectively, intersects the circles $ (A),\ (B),$ at their fixed points $ C,\ F$ ( easy to prove ).

Let $ D'E'$ be, the cord through the point $ P$ and parallel to the line segment $ AB,$ connecting the centers of the given circles $ (A),\ (B).$

So, we have that $ D'C\perp QE'$ and $ E'F\perp QD'$ and it is easy to show that the point $ H'\equiv D'C\cap E'F,$ as the orthocenter of the triangle $ \bigtriangleup QD'E',$ lies on $ PQ.$

Because of now, $ \angle CHQ \equal{} \angle DEQ \equal{} \angle D'E'Q \equal{} \angle CH'Q,$ we conclude that the point $ H\equiv DC\cap EF,$ as the orthocenter of the triangle $ \bigtriangleup QDE,$ lies on the fixed circle $ (K),$ taken as diameter the segment $ QH'$ and the solution is completed.

Kostas Vittas.


Show that $ i$ is a zero of $ P(x)\equal{}x^{3}\minus{}5x\plus{}6i$. Hence solve $ x^{3}\minus{}5x\plus{}6i\equal{}0$.

P(i)=0 hence (x-i) is a factor

$ x^{3}\minus{}5x\plus{}6i\equal{}(x\minus{}i)(x^{2}\plus{}bx\plus{}c)$

when x = 0,
$ 6i\equal{}\minus{}ci$
$ c\equal{}\minus{}6$

$ when x\equal{}1$,
$ \minus{}4\plus{}6i\equal{}(1\minus{}i)(\minus{}5\plus{}b)$
$ \minus{}4\plus{}6i\equal{}\minus{}5\plus{}b\plus{}5i\minus{}ib$
$ 5\plus{}b\equal{}4$
$ 5\minus{}b\equal{}6$

so $ b \equal{}\minus{}1$

upon expansion, it does not = P(x).

book has $ b\equal{}i$ :S

From $ \minus{}4\plus{}6i \equal{}\minus{}5\plus{}b\plus{}5i\minus{}ib$, you concluded $ 5\plus{}b \equal{} 4$ and $ 5\minus{}b \equal{} 6$ by taking the real and imaginary parts of each side. First, this only works when $ b$ is real, and since this polynomial is complex this may not be the case. Second, the first equation should actually be $ 5\minus{}b \equal{} 4$. So you've actually reached a contradiction by assuming $ b$ is real.

The correct thing to do is to simplify $ \minus{}4\plus{}6i \equal{}\minus{}5\plus{}b\plus{}5i\minus{}ib$. You'll get the following

$ 1\plus{}i \equal{} b(1\minus{}i)$

$ b \equal{} i$ is the unique solution to this.


m and n are chosen at random from the ${0, 1, 2,...10}$. What is the probability that $x^{2}+mx+n^{2}=0$ will have at least one real solution?

[quote="rofler"][hide] Discrimnant=b^2-4ac
=m^2-4n^2

m^2-4n^2 >= 0
m^2 >= (2n)^2
Since m and n are both non-negative
m >= 2n

m=2k has k+1 solutions
m=2k+1 has k+1 solutions

1+1+2+2+3+3+4+4+5+5+6 possiblilties
=36 possiblities

probability=36/100=9/25[/hide][/quote]

your denominator is wrong, other wise... its correct. $0-10$, there are $11$ numbers, so the denominator is $11^{2}=121$
should be $\frac{36}{121}$


number theory
relatively prime
superior algebra
superior algebra unsolved

For any intergers $ a,b$, show that $ \mathbb{Z}/(a\plus{}bi)$ always has exactly $ a^2\plus{}b^2$ elements, and if $ a, b$ are relatively prime, there is an isomorphism $ \mathbb{Z}/(a^2\plus{}b^2) \rightarrow \mathbb{Z}/(a\plus{}bi)$ but not otherwise.

If $ \mathbb Z/(a\plus{}bi)$ is substituted by $ \mathbb Z[i]/(a\plus{}bi)$, then it makes more sense to me. Do you mean that? ;)