2023 CMO Problems/Problem 5: Difference between revisions
Anyu tsuruko (talk | contribs) 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..." |
Anyu tsuruko (talk | contribs) No edit summary |
||
| (One intermediate revision by the same user not shown) | |||
| Line 1: | Line 1: | ||
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>. | == Problem == | ||
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 | ||
Latest revision as of 04:36, 25 May 2024
Problem
In an acute triangle 
is a point on the extension of 
. Through 
, draw lines parallel to 
and 
, denoted as 
and 
respectively, such that 
and 
. Let the circumcircle of 
intersect 
at point 
. Prove:
(1) 
;
(2) 
.
Solution 1
Proof for (1):
Let the side lengths of 
be 
. We have 
.
Let 
.
![\[\angle A=\angle B=\angle K^{\prime} \Rightarrow \frac{A B}{\sin C}=\frac{B C}{\sin A}=\frac{C A}{\sin B}\]](http://latex.artofproblemsolving.com/6/6/f/66fefee251ea330dfb59050372417b9384e5f341.png)
Assume :
![\[\begin{gathered} 2 K A \cos (\angle K-\angle A-\theta)=2 a \cos \theta \\ \Rightarrow \cos (\angle \theta)=2 a \cos (\angle C+\theta) \end{gathered}\]](http://latex.artofproblemsolving.com/c/e/f/cef904609c1a6e9ce99e27912890062e50fbdfd2.png)
Assume 
:
![\[\begin{gathered} \Rightarrow \angle B K C+\angle C=\theta, \quad \angle A^{\prime}=180^{\circ}-\angle C^{\prime}-(\angle A+B)=\angle K+A \\ \Rightarrow \frac{\sin (\angle A)}{\sin (\angle A+\theta)}=\frac{\cos B}{\cos C} \\ \Rightarrow \cos (\angle A-B) \leq \cos \theta \Rightarrow \cos (\angle A-B-C) \Rightarrow \cos (\angle A-B-C) \end{gathered}\]](http://latex.artofproblemsolving.com/a/9/0/a90f61837f89ff4cf479d3744b6d712ed37de63e.png)
Assume 
![\[A P: B T: C T=A B: C T: B P\]](http://latex.artofproblemsolving.com/0/a/4/0a4c14806ef7b461b497a4adac472f364aecbb63.png)
Proof: (2)
![\[\begin{gathered} A P=A R, A P=A S, C Q=C K, B P=B K \Rightarrow \frac{B T}{A R} \cdot \frac{A R}{C T} \cdot \frac{C A}{B P}=1 \Rightarrow(*) \\ \frac{B T}{A R}=\frac{\sin \angle B A T}{\sin \angle A B C}=\frac{\sin \angle A T}{\sin B} \Rightarrow \frac{A S}{C T}=\frac{\sin C}{\sin \theta} \end{gathered}\]](http://latex.artofproblemsolving.com/e/0/a/e0a51496cb8e28ffd81d4f7af8094dcc3b191094.png)
![\[CK=b \sin \theta \ (\sin (c-\theta), BK=c \sin (A+\theta) / \sin (c-\theta)\]](http://latex.artofproblemsolving.com/7/8/3/7834a5ac0d158e4e9f4eac31cadb20e30d8461b2.png)
![\[(*) \Leftrightarrow \frac{\sin (A+\theta)}{\sin B} \cdot \frac{\sin c}{\sin \theta} \cdot \frac{6 \sin \theta}{c \sin (A+\theta)}=1 \Leftrightarrow \frac{b}{c}=\frac{\sin B}{\sin C}\]](http://latex.artofproblemsolving.com/1/3/8/13894a9903e20aeba688411f03b24a085600fe7e.png)
~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 | ||
