It's a bit unexpected to see conics appearing on China TST again. Let $O, K$ be the circumcenter and symmedian point of $\triangle {ABC}$. The Kiepert Hyperbola ${kp}$ of $\triangle {ABC}$ is the isogonal image of the Brocard axis $OK$. It's well known that ${T}$ is on ${kp}$. ${M}$ is the midpoint of $AB$. By the angle conditions, ${K}$ lies in the interior of $\triangle {MBO}$, so line $OK$ intersects segments $BA, BC$. Therefore, $A, C$ are on the same branch of ${kp}$. However, if $\angle BTC$ is right, the tangent of ${T}$ to ${kp}$ is perpendicular to $BC$, and it intersects segment $BC$. This means that separated by a tangent, $B, C$ are on different branches of ${kp}$, which is absurd.

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so what am i doing wrong

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DottedCaculator wrote: so what am i doing wrong Sorry I didn't type the problem correctly. It should be ATC instead of BTC.

I wanted to find a synthetic proof, but ended up with using hyperbolas. We use 3 well known properties of kiepert hyperbolas Lemma 1: Brocard axis is isogonal conjugate of Kiepert hyberbola Lemma 2: If $A,B.C$ lie on a rectangular hyperbola s.t. $\measuredangle ABC = 90^{\circ}$ then the tangent at $B$ is perpendicular to $AC$ Lemma 3: Kiepert hyperbola is rectangular hyperbola. Claim 1: If $A,B,C$ lie on same branch of hyperbola then $\measuredangle ABC \neq 90^{\circ}$

Proof

Let $K$ be the symmedian point. Since $OK$ intersects $AB,AC$ (this is true by angles)) so $A,C$ belong to the same branch as $T$, now by Claim 1, we're done

Finally done, first TST2 Geo

Define $A'$ similarly. By Jacobi, $AA'$, $BB'$, and $CC'$ concur. Swap $A$ and $B$ since I do not want to retype this entire complex bash again. Let $|\omega|=1$, $y=-\frac1{\omega-1}$, and $z=\frac{\omega}{\omega-1}$. We get $b_1=yc+za$ and $c_1=ya+zb$, so $\angle BTC=90^\circ$ if and only if \begin{align*} \frac{ya+zb-c}{yc+za-b}&=-\frac{\overline y\frac1a+\overline z\frac1b-\frac1c}{\overline y\frac1c+\overline z\frac1a-\frac1b}\\ \frac{-a+\omega b+(1-\omega)c}{-c+\omega a+(1-\omega)b}&=-\frac{-\omega bc+ca+(\omega-1)ab}{-\omega ab+bc+(\omega-1)ca}\\ (-a+\omega b+(1-\omega)c)(-\omega ab+bc+(\omega-1)ca)&=-(-c+\omega a+(1-\omega)b)(-\omega bc+ca+(\omega-1)ab)\\ \omega a^2b+(1-\omega)a^2c-\omega^2 ab^2+\omega b^2c+(1-\omega)bc^2-(1-\omega)^2ac^2+(-1+2\omega^2-2\omega)abc&=\omega(1-\omega)a^2b-\omega a^2c+(1-\omega)^2ab^2+\omega(1-\omega)b^2c-\omega bc^2+ac^2+(2\omega-2+\omega^2)abc\\ \omega^2 a^2b+a^2c+(-2\omega^2+2\omega-1)ab^2+\omega^2 b^2c+bc^2+(-\omega^2+2\omega-2)ac^2+(\omega^2-4\omega+1)abc&=0\\ \omega^2(a^2b-2ab^2+b^2c-ac^2+abc)+\omega(2ab^2+2ac^2-4abc)+a^2c-ab^2+bc^2-2ac^2+abc&=0\\ \frac bc+\frac cb-2&=\Re\left(\omega\left(\frac ac-2\frac bc+\frac ba-\frac cb+1\right)\right). \end{align*} We need to show $2-\frac bc-\frac cb>\left|\frac ac-2\frac bc+\frac ba-\frac cb+1\right|$ when $\angle B>\angle A>\angle C$. WLOG $c=1$ and $\Im(b)>0$, so $\Im(a)<0$. $\angle A>\angle C$ implies $\Re(b)<0$. We need to show $2-b-\frac1b>|a-2b+b/a-1/b+1|$. The minimum of the right hand side occurs at $a=-1$ or $a=b^2$. In the first case, the right hand side is $|3b^2+1|$, so we need to show $10+3b^2+\frac3{b^2}<b^2-4b+6-\frac4b+\frac1{b^2}$, or $2b^2+4b+4+\frac4b+\frac2{b^2}=2(b+2+1/b)(b+1/b)<0$, which is true since the first term is positive and the second term is negative. In the second case, the right hand side is $|b^2-2b+1|=2-b-\frac1b$. Therefore, the inequality has no solutions, so $\angle BTC$ cannot be a right angle.

三！角！法！好！

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Why must T lie on the Kiepert Hyperbola?

sami1618 wrote: Why must T lie on the Kiepert Hyperbola? In fact, it's a well-known nature of Kiepert Hyperbola. See details at https://tieba.baidu.com/p/6501803387 (I‘m sorry it's written by Chinese...)

zhihanpeng wrote: 三！角！法！好！ 膜爆pengle 膜爆pengle 膜爆pengle 膜爆pengle 膜爆pengle 膜爆pengle 膜爆pengle 膜爆pengle 膜爆pengle 膜爆pengle