Difference between revisions of "2023 CMO Problems/Problem 5"
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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>. | + | 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 04:34, 25 May 2024
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 .
Assume :
Assume :
Assume Proof: (2)
~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 |