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# AnhTran2911

Đăng ký: 10-03-2017
Offline Đăng nhập: Hôm qua, 16:14

### Thử latex

12-09-2018 - 21:27

3) Ta chọn được một số $n_{0}$ đủ lớn thỏa mãn mỗi $2017^{n_{0}}+2018^{n_{0}}-i$ có một ước nguyên tố $p_{i}$, các số nguyên tố này không nhất thiết phân biệt ( Với $i$ chạy từ $1$ đến $k$)

Lúc này để chứng minh dãy trên toàn hợp số với vô hạn $n$ thì ta chỉ cần chọn $n=n_{0}+t.\Pi_{1}^k(p_i-1)$ và từ đó theo định lí $Fermat$ bé ta có $2017^n+2018^n-i$ chia hết cho $p_{i}$ nên dãy này toàn là hợp số với vô số $n$ do ta cho $t$ chạy ra vô hạn

### Ko73 cubic

25-08-2018 - 17:31

Chứng minh rằng điểm Kosnita của một tam giác nằm trên K073 cubic của tam giác đó

### Related to point lies on Kiepert hyperbola

21-08-2018 - 23:52

Problem : If $P$ is a point on the Kiepert hyperbola then the circumcenter of its anticevian triangle, the orthocenter of $\triangle ABC$ and $P$ are collinear.
Solution : TelvCohl from AOPS

Lemma 1 : Given a $\triangle ABC$ and a point $P$. Let $\triangle A^*B^*C^*$ be the circumcevian triangle of $P$ WRT $\triangle ABC$. Let $K,$ $K^*$ be the symmedian point of $\triangle ABC,$ $\triangle A^*B^*C^*$, respectively. Then $K,$ $P,$ $K^*$ are collinear.

Proof : Let $T$ $\equiv$ $AK$ $\cap$ $\odot (ABC)$ and $T^*$ $\equiv$ $A^*K^*$ $\cap$ $\odot (A^*B^*C^*)$. Let the Lemoine axis of $\triangle ABC$ cuts $BC,$ $CA,$ $AB$ at $D,$ $E,$ $F$, respectively and let the Lemoine axis of $\triangle A^*B^*C^*$ cuts $B^*C^*,$ $C^*A^*,$ $A^*B^*$ at $D^*,$ $E^*,$ $F^*$, respectively. Let $A_1$ $\equiv$ $BC$ $\cap$ $B^*C^*,$ $B_1$ $\equiv$ $CA$ $\cap$ $C^*A^*,$ $C_1$ $\equiv$ $AB$ $\cap$ $A^*B^*$ and let $O$ be the circumcenter of $\triangle ABC$ ($\triangle A^*B^*C^*$).

Clearly, $A_1,$ $B_1,$ $C_1$ lie on the polar $\tau$ of $P$ WRT $\odot (O)$. Since $ABTC$ and $A^*B^*T^*C^*$ are harmonic quadrilateral, so $T,$ $P,$ $T^*$ are collinear $\Longrightarrow$ $AK$ $\cap$ $A^*K^*$ $\in$ $\tau$. Similarly, we can prove $BK$ $\cap$ $B^*K^*$ $\in$ $\tau$ and $CK$ $\cap$ $C^*K^*$ $\in$ $\tau$. Since the tangent of $\odot (O)$ through $B,$ $C$ and $AK$ are concurrent, so $D$ lies on the polar of $AK$ $\cap$ $A^*K^*$ WRT $\odot (O)$. Similarly, we can prove $D^*$ lies on the polar of $AK$ $\cap$ $A^*K^*$ WRT $\odot (O)$ $\Longrightarrow$ $D,$ $P,$ $D^*$ are collinear. Analogously, we can prove $P$ $\in$ $EE^*$ and $P$ $\in$ $FF^*$, so from Desargue theorem ($\triangle B_1EE^*$ and $\triangle C_1FF^*$) we get $\tau,$ $EF,$ $E^*F^*$ are concurrent, hence their pole $P,$ $K,$ $K^*$ WRT $\odot (O)$ are collinear.
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Lemma 2 : Let $P$ be a point on the Kiepert hyperbola of $\triangle ABC$. Let $\triangle P_aP_bP_c$ be the pedal triangle of $P$ WRT $\triangle ABC$ and let $\triangle XYZ$ be the circumcevian triangle of $P$ WRT $\triangle P_aP_bP_c$. Then $P$ lies on the Kiepert hyperbola of $\triangle XYZ$.

Proof : Let $Q$ be the isogonal conjugate of $P$ WRT $\triangle ABC$. Let $\triangle Q_aQ_bQ_c,$ $\triangle Q_AQ_BQ_C$ be the pedal triangle, circumcevian triangle of $Q$ WRT $\triangle ABC$. Let $R$ be the isogonal conjugate of $Q$ WRT $\triangle Q_AQ_BQ_C$. Since $\triangle Q_AQ_BQ_C$ $\cup$ $R$ $\sim$ $\triangle Q_aQ_bQ_c$ $\cup$ $Q$ $\cong$ $\triangle XYZ$ $\cup$ $P$, so it suffices to prove $Q$ lies on the Brocard axis of $\triangle Q_AQ_BQ_C$. Let $O$ be the circumcenter of $\triangle ABC$. Let $K,$ $K_Q$ be the symmedian point of $\triangle ABC,$ $\triangle Q_AQ_BQ_C$, respectively. From Lemma 1 we get $K,$ $Q,$ $K_Q$ are collinear, so notice $Q$ lies on the Brocard axis $OK$ of $\triangle ABC$ we conclude that $Q$ $\in$ $OK_Q$ (Brocard axis of $\triangle Q_AQ_BQ_C$).

Remark : There is a stronger result of Lemma 2 : If $P$ is the Kiepert perspector of $\triangle ABC$ with angle $\theta$, then $P$ is the Kiepert perspector of $\triangle XYZ$ with angle $-\theta$ (but we don't need this stronger result in the proof).
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Now we recall two well-known properties about conic as following :

Property 1 : Given a $\triangle ABC$ and two points $P,$ $Q$. Let $\triangle P_aP_bP_c$ be the anticevian triangle of $P$ WRT $\triangle ABC$ and let $\triangle Q_aQ_bQ_c$ be the anticevian triangle of $Q$ WRT $\triangle ABC$. Then $P,$ $Q,$ $P_a,$ $P_b,$ $P_c,$ $Q_a,$ $Q_b,$ $Q_c$ lie on a conic.

Property 2 : Given a $\triangle ABC$ and a point $P$. Let $I,$ $I_a,$ $I_b,$ $I_c$ be the incenter, A-excenter, B-excenter, C-excenter of $\triangle ABC$, respectively. Let $\mathcal{H}$ be a conic passing through $I,$ $I_a,$ $I_b,$ $I_c$. Then the polar of $P$ WRT $\mathcal{H}$ passes through $P^*$ where $P^*$ is the isogonal conjugate of $P$ WRT $\triangle ABC$.
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From Property 1 and Property 2 we get the following lemma :

Lemma 3 : Let $\mathcal{H}$ be a circum-rectangular hyperbola of $\triangle ABC$ and let $P,$ $Q$ be the points on $\mathcal{H}$. Let $\triangle DEF$ be the cevian triangle of $Q$ WRT $\triangle ABC$ and let $P^*$ be the isogonal conjugate of $P$ WRT $\triangle DEF$. Then $PP^*$ is tangent to $\mathcal{H}$.

Proof : Let $I,$ $I_a,$ $I_b,$ $I_c$ be the incenter, A-excenter, B-excenter, C-excenter of $\triangle DEF$, respectively. From Property 1 we get $A,$ $B,$ $C,$ $Q,$ $I,$ $I_a,$ $I_b,$ $I_c$ lie on a conic $\mathcal{C}$, but notice $I$ is the orthocenter of $\triangle I_aI_bI_c$ we get $\mathcal{C}$ is a rectangular hyperbola $\Longrightarrow$ $\mathcal{C}$ $\equiv$ $\mathcal{H}$, so from Property 2 we conclude that $P^*$ lies on the polar of $P$ WRT $\mathcal{H}$. i.e. $PP^*$ is tangent to $\mathcal{H}$
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Back to the main problem :

Let $H$ be the orthocenter of $\triangle ABC$. Let $\triangle XYZ$ be the anticevian triangle of $P$ WRT $\triangle ABC$ and $J$ be the circumcenter of $\triangle XYZ$. Let $\triangle A_1B_1C_1$ be the pedal triangle of $P$ WRT $\triangle ABC$. Perform the Inversion with center $P$ and denote $V^*$ as the image of $V$ ($V$ is an arbitrary point). Obviously, $\triangle A^*B^*C^*$ is the pedal triangle of $P$ WRT $\triangle A_1^*B_1^*C_1^*$ and $\triangle X^*Y^*Z^*$ is the pedal triangle of $P$ WRT the medial triangle $\triangle A_2^*B_2^*C_2^*$ of $\triangle A_1^*B_1^*C_1^*$, so $PJ$ $\equiv$ $PJ^*$ passes through the isogonal conjugate $Q^*$ of $P$ WRT $\triangle A_2^*B_2^*C_2^*$.

Let $\triangle DEF$ be the anticomplementary triangle of $\triangle ABC$. Let $\triangle A_3^*B_3^*C_3^*$ be the cevian triangle of $P$ WRT $\triangle A_1^*B_1^*C_1^*$. From $EF$ $\parallel$ $BC$ and $A$ $\in$ $EF$ $\Longrightarrow$ the image of the line $EF$ under the Inversion is the circle with diameter $PA_3^*$. Similarly, we can prove $\odot (PB_3^*),$ $\odot (PC_3^*)$ is the image of the line $FD,$ $DE$ under the Inversion, respectively, so $\triangle D^*E^*F^*$ is the pedal triangle of $P$ WRT $\triangle A_3^*B_3^*C_3^*$. Since $H$ is the circumcenter of $\triangle DEF$, so we get $PH$ $\equiv$ $PH^*$ passes through the isogonal conjugate $R^*$ of $P$ WRT $\triangle A_3^*B_3^*C_3^*$.

Let $G^*$ $\equiv$ $A_1^*A_2^*$ $\cap$ $B_1^*B_2^*$ $\cap$ $C_1^*C_2^*$ be the Centroid of $\triangle A_1^*B_1^*C_1^*$. From Lemma 2 we know $P$ lies on the Kiepert hyperbola of $\triangle A_1^*B_1^*C_1^*$, so $A_1^*,$ $B_1^*,$ $C_1^*,$ $G^*,$ $P$ lie on a rectangular hyperbola $\mathcal{K}$, hence from Lemma 3 we conclude that $P,$ $Q^*,$ $R^*$ lie on the tangent of $\mathcal{K}$ through $P$ $\Longrightarrow$ $P,$ $J^*,$ $H^*$ are collinear $\Longrightarrow$ $P,$ $J,$ $H$ are collinear.

### $(x+1)(y+1)\mid{x^3+y^3+1}$

11-11-2017 - 22:29

Tìm $x,y\in{Z^+}$ thỏa mãn $(x+1)(y+1)\mid{x^3+y^3+1}.$

Spoiler