Track / Stage 6 / 230 of 400 #1230 of 2000
Problem 1230 National olympiad, first round Difficulty 6.4 Prove it
[ Second-order curves ]
Prove that under the rotation x ′ ′ = x ′ cos φ + y ′ sin φ , y ′ ′ = − x ′ sin φ + y ′ cos φ x^{\prime \prime}=x^{\prime} \cos \varphi+y^{\prime} \sin \varphi, y^{\prime \prime}=-x^{\prime} \sin \varphi+y^{\prime} \cos \varphi x ′′ = x ′ cos φ + y ′ sin φ , y ′′ = − x ′ sin φ + y ′ cos φ a x ′ 2 + 2 b x ′ y ′ + c y ′ 2 a x^{\prime 2}+2 b x^{\prime} y^{\prime}+c y^{\prime 2} a x ′2 + 2 b x ′ y ′ + c y ′2 a 1 x ′ 2 + 2 b 1 x ′ ′ y ′ ′ + c 1 y ′ 2 a_{1} x^{\prime 2}+2 b_{1} x^{\prime \prime} y^{\prime \prime}+c_{1} y^{\prime 2} a 1 x ′2 + 2 b 1 x ′′ y ′′ + c 1 y ′2 a 1 c 1 − b 1 2 = a c − b 2 a_{1} c_{1}-b_{1}^{2}=a c-b^{2} a 1 c 1 − b 1 2 = a c − b 2
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Official solution When solving problem 31.002 ‾ \underline{31.002} 31.002
a 1 = a cos 2 φ − 2 b cos φ sin φ + c sin 2 φ b 1 = a cos φ sin φ + b ( cos 2 φ − sin 2 φ ) − c cos φ sin φ c 1 = a sin 2 φ + 2 b cos φ sin φ + c cos 2 φ
\begin{aligned}
& a_{1}=a \cos ^{2} \varphi-2 b \cos \varphi \sin \varphi+c \sin ^{2} \varphi \\
& b_{1}=a \cos \varphi \sin \varphi+b\left(\cos ^{2} \varphi-\sin ^{2} \varphi\right)-c \cos \varphi \sin \varphi \\
& c_{1}=a \sin ^{2} \varphi+2 b \cos \varphi \sin \varphi+c \cos ^{2} \varphi
\end{aligned}
a 1 = a cos 2 φ − 2 b cos φ sin φ + c sin 2 φ b 1 = a cos φ sin φ + b ( cos 2 φ − sin 2 φ ) − c cos φ sin φ c 1 = a sin 2 φ + 2 b cos φ sin φ + c cos 2 φ
Therefore,
a 1 c 1 − b 1 2 = ( a + c ) sin 2 φ cos 2 φ + a c ( sin 4 φ + cos 4 φ ) − − 2 b ( a − c ) sin φ cos φ ( sin 2 φ − cos 2 φ ) − 4 b 2 sin 2 φ cos 2 φ − − ( a + c ) sin 2 φ cos 2 φ + 2 a c sin 2 φ cos 2 φ − − 2 b ( a − c ) sin φ cos φ ( cos 2 φ − sin 2 φ ) − b 2 ( cos 2 φ − sin 2 φ ) 2 = = a c − b 2
\begin{aligned}
a_{1} c_{1}-b_{1}{ }^{2} & =(a+c) \sin ^{2} \varphi \cos ^{2} \varphi+a c\left(\sin ^{4} \varphi+\cos ^{4} \varphi\right)- \\
& -2 b(a-c) \sin \varphi \cos \varphi\left(\sin ^{2} \varphi-\cos ^{2} \varphi\right)-4 b^{2} \sin ^{2} \varphi \cos ^{2} \varphi- \\
& -(a+c) \sin ^{2} \varphi \cos ^{2} \varphi+2 a c \sin ^{2} \varphi \cos ^{2} \varphi- \\
& -2 b(a-c) \sin \varphi \cos \varphi\left(\cos ^{2} \varphi-\sin ^{2} \varphi\right)-b^{2}\left(\cos ^{2} \varphi-\sin ^{2} \varphi\right)^{2}= \\
& =a c-b^{2}
\end{aligned}
a 1 c 1 − b 1 2 = ( a + c ) sin 2 φ cos 2 φ + a c ( sin 4 φ + cos 4 φ ) − − 2 b ( a − c ) sin φ cos φ ( sin 2 φ − cos 2 φ ) − 4 b 2 sin 2 φ cos 2 φ − − ( a + c ) sin 2 φ cos 2 φ + 2 a c sin 2 φ cos 2 φ − − 2 b ( a − c ) sin φ cos φ ( cos 2 φ − sin 2 φ ) − b 2 ( cos 2 φ − sin 2 φ ) 2 = = a c − b 2
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