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Difference between revisions of "2023 CMO Problems/Problem 5"

(Created page with "In an acute triangle <math>\triangle A B C, K</math> is a point on the extension of <math>B C</math>. Through <math>K</math>, draw lines parallel to <math>A B</math> and <math...")
 
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In an acute triangle <math>\triangle A B C, K</math> is a point on the extension of <math>B C</math>. Through <math>K</math>, draw lines parallel to <math>A B</math> and <math>A C</math>, denoted as <math>K P</math> and <math>K Q</math> respectively, such that <math>B K=B P</math> and <math>C K=C Q</math>. Let the circumcircle of <math>\triangle K P Q</math> intersect <math>A K</math> at point <math>T</math>. Prove: (1) <math>\angle B T C+\angle A P B=\angle C Q A</math>; (2) <math>A P \cdot B T \cdot C Q=A Q \cdot C T \cdot B P</math>.
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In an acute triangle <math>\triangle A B C, K</math> is a point on the extension of <math>B C</math>. Through <math>K</math>, draw lines parallel to <math>A B</math> and <math>A C</math>, denoted as <math>K P</math> and <math>K Q</math> respectively, such that <math>B K=B P</math> and <math>C K=C Q</math>. Let the circumcircle of <math>\triangle K P Q</math> intersect <math>A K</math> at point <math>T</math>. Prove:  
 +
 
 +
(1) <math>\angle B T C+\angle A P B=\angle C Q A</math>;  
 +
 
 +
(2) <math>A P \cdot B T \cdot C Q=A Q \cdot C T \cdot B P</math>.
 
==Solution 1==
 
==Solution 1==
 +
Proof for (1):
 +
Let the side lengths of <math>\triangle A B C</math> be <math>a, b, c</math>. We have <math>K A=K^{\prime} T-2 K B-K C</math>.
 +
Let <math>B C=a, C A=b, A B=c</math>.
 +
<cmath>
 +
\angle A=\angle B=\angle K^{\prime} \Rightarrow \frac{A B}{\sin C}=\frac{B C}{\sin A}=\frac{C A}{\sin B}
 +
</cmath>
 +
 +
Assume <math>K A=K^{\prime} T-2 K B-K C</math> :
 +
<cmath>
 +
\begin{gathered}
 +
2 K A \cos (\angle K-\angle A-\theta)=2 a \cos \theta \\
 +
\Rightarrow \cos (\angle \theta)=2 a \cos (\angle C+\theta)
 +
\end{gathered}
 +
</cmath>
 +
 +
Assume <math>K T \sin A=K P \sin \theta+C \sin (\angle A+\theta)</math> :
 +
<cmath>
 +
\begin{gathered}
 +
\Rightarrow \angle B K C+\angle C=\theta, \quad \angle A^{\prime}=180^{\circ}-\angle C^{\prime}-(\angle A+B)=\angle K+A \\
 +
\Rightarrow \frac{\sin (\angle A)}{\sin (\angle A+\theta)}=\frac{\cos B}{\cos C} \\
 +
\Rightarrow \cos (\angle A-B) \leq \cos \theta \Rightarrow \cos (\angle A-B-C) \Rightarrow \cos (\angle A-B-C)
 +
\end{gathered}
 +
</cmath>
  
 +
Assume <math>\frac{a}{\sin B}=\frac{b}{\sin A}</math>
 +
<cmath>
 +
A P: B T: C T=A B: C T: B P
 +
</cmath>
 +
Proof: (2)
 +
<cmath>
 +
\begin{gathered}
 +
A P=A R, A P=A S, C Q=C K, B P=B K \Rightarrow \frac{B T}{A R} \cdot \frac{A R}{C T} \cdot \frac{C A}{B P}=1 \Rightarrow(*) \\
 +
\frac{B T}{A R}=\frac{\sin \angle B A T}{\sin \angle A B C}=\frac{\sin \angle A T}{\sin B} \Rightarrow \frac{A S}{C T}=\frac{\sin C}{\sin \theta}
 +
\end{gathered}
 +
</cmath>
 +
<cmath>
 +
CK=b \sin \theta \ (\sin (c-\theta), BK=c \sin (A+\theta) / \sin (c-\theta)
 +
</cmath>
 +
<cmath>
 +
(*) \Leftrightarrow \frac{\sin (A+\theta)}{\sin B} \cdot \frac{\sin c}{\sin \theta} \cdot \frac{6 \sin \theta}{c \sin (A+\theta)}=1 \Leftrightarrow \frac{b}{c}=\frac{\sin B}{\sin C}
 +
</cmath>
  
 
~xiaohuangya|szm
 
~xiaohuangya|szm

Revision as of 05:34, 25 May 2024

In an acute triangle $\triangle A B C, K$ is a point on the extension of $B C$. Through $K$, draw lines parallel to $A B$ and $A C$, denoted as $K P$ and $K Q$ respectively, such that $B K=B P$ and $C K=C Q$. Let the circumcircle of $\triangle K P Q$ intersect $A K$ at point $T$. Prove:

(1) $\angle B T C+\angle A P B=\angle C Q A$;

(2) $A P \cdot B T \cdot C Q=A Q \cdot C T \cdot B P$.

Solution 1

Proof for (1): Let the side lengths of $\triangle A B C$ be $a, b, c$. We have $K A=K^{\prime} T-2 K B-K C$. Let $B C=a, C A=b, A B=c$. \[\angle A=\angle B=\angle K^{\prime} \Rightarrow \frac{A B}{\sin C}=\frac{B C}{\sin A}=\frac{C A}{\sin B}\]

Assume $K A=K^{\prime} T-2 K B-K C$ : \[\begin{gathered} 2 K A \cos (\angle K-\angle A-\theta)=2 a \cos \theta \\ \Rightarrow \cos (\angle \theta)=2 a \cos (\angle C+\theta) \end{gathered}\]

Assume $K T \sin A=K P \sin \theta+C \sin (\angle A+\theta)$ : \[\begin{gathered} \Rightarrow \angle B K C+\angle C=\theta, \quad \angle A^{\prime}=180^{\circ}-\angle C^{\prime}-(\angle A+B)=\angle K+A \\ \Rightarrow \frac{\sin (\angle A)}{\sin (\angle A+\theta)}=\frac{\cos B}{\cos C} \\ \Rightarrow \cos (\angle A-B) \leq \cos \theta \Rightarrow \cos (\angle A-B-C) \Rightarrow \cos (\angle A-B-C) \end{gathered}\]

Assume $\frac{a}{\sin B}=\frac{b}{\sin A}$ \[A P: B T: C T=A B: C T: B P\] Proof: (2) \[\begin{gathered} A P=A R, A P=A S, C Q=C K, B P=B K \Rightarrow \frac{B T}{A R} \cdot \frac{A R}{C T} \cdot \frac{C A}{B P}=1 \Rightarrow(*) \\ \frac{B T}{A R}=\frac{\sin \angle B A T}{\sin \angle A B C}=\frac{\sin \angle A T}{\sin B} \Rightarrow \frac{A S}{C T}=\frac{\sin C}{\sin \theta} \end{gathered}\] \[CK=b \sin \theta \ (\sin (c-\theta), BK=c \sin (A+\theta) / \sin (c-\theta)\] \[(*) \Leftrightarrow \frac{\sin (A+\theta)}{\sin B} \cdot \frac{\sin c}{\sin \theta} \cdot \frac{6 \sin \theta}{c \sin (A+\theta)}=1 \Leftrightarrow \frac{b}{c}=\frac{\sin B}{\sin C}\]

~xiaohuangya|szm

See Also

2023 CMO(CHINA) (Problems • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6
All CMO(CHINA) Problems and Solutions
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