An algebra problem by James Wilson

Algebra Level 3

Let x , y , a 1 , a 2 , b 1 , b 2 , c 1 , c 2 x,y,a_1,a_2,b_1,b_2,c_1,c_2 be real numbers.

Given

a 1 ( x 2 − y 2 ) − 2 a 2 x y + b 1 x − b 2 y + c 1 = 0 a_1(x^2-y^2)-2a_2xy+b_1x-b_2y+c_1=0 ,

a 2 ( x 2 − y 2 ) + 2 a 1 x y + b 2 x + b 1 y + c 2 = 0 a_2(x^2-y^2)+2a_1xy+b_2x+b_1y+c_2=0 ,

b 1 b 2 = 2 ( a 1 c 2 + a 2 c 1 ) b_1b_2=2(a_1c_2+a_2c_1) ,

and

b 1 2 − b 2 2 = 4 ( a 1 c 1 − a 2 c 2 ) b_1^2-b_2^2=4(a_1c_1-a_2c_2) ,

y y takes the form y = 1 m a n b p − a q b r a s t + a u v y=\frac{1}{m}\frac{a_nb_p-a_qb_r}{a_s^t+a_u^v} .

Find m 2 − n 2 + p 2 − q 2 + r 2 − s 2 + t 2 − u 2 + v 2 m^2-n^2+p^2-q^2+r^2-s^2+t^2-u^2+v^2 .

7 -7 -3 2

James Wilson by James Wilson , 28, USA

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1 solution

James Wilson
Nov 8, 2020

Multiply the second equation by i i ; then add the first two equations; then proceed as follows: a 1 ( x 2 − y 2 ) − 2 a 2 x y + b 1 x − b 2 y + c 1 + i [ a 2 ( x 2 − y 2 ) + 2 a 1 x y + b 2 x + b 1 y + c 2 ] = 0 a_1(x^2-y^2)-2a_2xy+b_1x-b_2y+c_1+i[a_2(x^2-y^2)+2a_1xy+b_2x+b_1y+c_2]=0 ⇔ a 1 ( x 2 − y 2 ) − 2 a 2 x y + b 1 x − b 2 y + c 1 + i a 2 ( x 2 − y 2 ) + i 2 a 1 x y + i b 2 x + i b 1 y + i c 2 = 0 \Leftrightarrow a_1(x^2-y^2)-2a_2xy+b_1x-b_2y+c_1+ia_2(x^2-y^2)+i2a_1xy+ib_2x+ib_1y+ic_2=0 ⇔ a 1 ( x 2 − y 2 ) + i a 2 ( x 2 − y 2 ) + 2 i a 1 x y − 2 a 2 x y + b 1 x + i b 2 x + i b 1 y − b 2 y + c 1 + i c 2 = 0 \Leftrightarrow a_1(x^2-y^2)+ia_2(x^2-y^2)+2ia_1xy-2a_2xy+b_1x+ib_2x+ib_1y-b_2y+c_1+ic_2=0 ⇔ ( a 1 + i a 2 ) ( x 2 − y 2 ) + ( i a 1 − a 2 ) ( 2 x y ) + ( b 1 + i b 2 ) x + ( i b 1 − b 2 ) y + c 1 + i c 2 = 0 \Leftrightarrow (a_1+ia_2)(x^2-y^2)+(ia_1-a_2)(2xy)+(b_1+ib_2)x+(ib_1-b_2)y+c_1+ic_2=0 ⇔ ( a 1 + i a 2 ) ( x 2 − y 2 ) + ( i ) ( − i ) ( i a 1 − a 2 ) ( 2 x y ) + ( b 1 + i b 2 ) x + ( i ) ( − i ) ( i b 1 − b 2 ) y + c 1 + i c 2 = 0 \Leftrightarrow (a_1+ia_2)(x^2-y^2)+(i)(-i)(ia_1-a_2)(2xy)+(b_1+ib_2)x+(i)(-i)(ib_1-b_2)y+c_1+ic_2=0 ⇔ ( a 1 + i a 2 ) ( x 2 − y 2 ) + ( i ) ( − i 2 a 1 − ( − i ) a 2 ) ( 2 x y ) + ( b 1 + i b 2 ) x + ( i ) ( − i 2 b 1 − ( − i ) b 2 ) y + c 1 + i c 2 = 0 \Leftrightarrow (a_1+ia_2)(x^2-y^2)+(i)(-i^2a_1-(-i)a_2)(2xy)+(b_1+ib_2)x+(i)(-i^2b_1-(-i)b_2)y+c_1+ic_2=0 ⇔ ( a 1 + i a 2 ) ( x 2 − y 2 ) + i ( a 1 + i a 2 ) ( 2 x y ) + ( b 1 + i b 2 ) x + i ( b 1 + b 2 i ) y + c 1 + i c 2 = 0 \Leftrightarrow (a_1+ia_2)(x^2-y^2)+i(a_1+ia_2)(2xy)+(b_1+ib_2)x+i(b_1+b_2i)y+c_1+ic_2=0 ⇔ ( a 1 + i a 2 ) ( x 2 − y 2 ) + ( a 1 + i a 2 ) ( 2 i x y ) + ( b 1 + i b 2 ) x + ( b 1 + b 2 i ) i y + c 1 + i c 2 = 0 \Leftrightarrow (a_1+ia_2)(x^2-y^2)+(a_1+ia_2)(2ixy)+(b_1+ib_2)x+(b_1+b_2i)iy+c_1+ic_2=0 ⇔ ( a 1 + i a 2 ) ( x 2 − y 2 + 2 i x y ) + ( b 1 + i b 2 ) ( x + i y ) + c 1 + i c 2 = 0 \Leftrightarrow (a_1+ia_2)(x^2-y^2+2ixy)+(b_1+ib_2)(x+iy)+c_1+ic_2=0 ⇔ ( a 1 + i a 2 ) ( x 2 + i x y + i x y + i 2 y 2 ) + ( b 1 + i b 2 ) ( x + i y ) + c 1 + i c 2 = 0 \Leftrightarrow (a_1+ia_2)(x^2+ixy+ixy+i^2y^2)+(b_1+ib_2)(x+iy)+c_1+ic_2=0 ⇔ ( a 1 + i a 2 ) ( x ( x + i y ) + i y ( x + i y ) ) + ( b 1 + i b 2 ) ( x + i y ) + c 1 + i c 2 = 0 \Leftrightarrow (a_1+ia_2)(x(x+iy)+iy(x+iy))+(b_1+ib_2)(x+iy)+c_1+ic_2=0 ⇔ ( a 1 + i a 2 ) ( x + i y ) 2 + ( b 1 + i b 2 ) ( x + i y ) + c 1 + i c 2 = 0 \Leftrightarrow (a_1+ia_2)(x+iy)^2+(b_1+ib_2)(x+iy)+c_1+ic_2=0 For simplicity, let z = x + i y , A = a 1 + i a 2 , B = b 1 + i b 2 , C = c 1 + i c 2 z=x+iy,A=a_1+ia_2,B=b_1+ib_2, C=c_1+ic_2 The equation then becomes A z 2 + B z + C = 0 Az^2+Bz+C=0 ⇔ A ( z ) ( z + B A ) + C = 0 \Leftrightarrow A(z)\Big(z+\frac{B}{A}\Big)+C=0 ⇔ ( z ) ( z + B A ) + C A = 0 \Leftrightarrow (z)\Big(z+\frac{B}{A}\Big)+\frac{C}{A}=0 ⇔ ( z + ( B 2 A − B 2 A ) ) ( z + ( B 2 A + B 2 A ) ) + C A = 0 \Leftrightarrow \Big(z+\Big(\frac{B}{2A}-\frac{B}{2A}\Big)\Big)\Big(z+\Big(\frac{B}{2A}+\frac{B}{2A}\Big)\Big)+\frac{C}{A}=0 ⇔ ( ( z + B 2 A ) − B 2 A ) ( ( z + B 2 A ) + B 2 A ) + C A = 0 \Leftrightarrow \Big(\Big(z+\frac{B}{2A}\Big)-\frac{B}{2A}\Big)\Big(\Big(z+\frac{B}{2A}\Big)+\frac{B}{2A}\Big)+\frac{C}{A}=0 ⇔ ( z + B 2 A ) 2 + ( z + B 2 A ) B 2 A − B 2 A ( z + B 2 A ) − ( B 2 A ) 2 + C A = 0 \Leftrightarrow \Big(z+\frac{B}{2A}\Big)^2+\Big(z+\frac{B}{2A}\Big)\frac{B}{2A}-\frac{B}{2A}\Big(z+\frac{B}{2A}\Big)-\Big(\frac{B}{2A}\Big)^2+\frac{C}{A}=0 ⇔ ( z + B 2 A ) 2 − ( B 2 A ) 2 + C A = 0 \Leftrightarrow \Big(z+\frac{B}{2A}\Big)^2-\Big(\frac{B}{2A}\Big)^2+\frac{C}{A}=0 ⇔ ( z + B 2 A ) 2 = ( B 2 A ) 2 − C A \Leftrightarrow \Big(z+\frac{B}{2A}\Big)^2=\Big(\frac{B}{2A}\Big)^2-\frac{C}{A} ⇔ ( z + B 2 A ) 2 = B 2 4 A 2 − C A ⋅ 4 A 4 A \Leftrightarrow \Big(z+\frac{B}{2A}\Big)^2=\frac{B^2}{4A^2}-\frac{C}{A}\cdot\frac{4A}{4A} ⇔ ( z + B 2 A ) 2 = B 2 − 4 A C 4 A 2 \Leftrightarrow \Big(z+\frac{B}{2A}\Big)^2=\frac{B^2-4AC}{4A^2} ⇔ z + B 2 A = ± B 2 − 4 A C 2 A \Leftrightarrow z+\frac{B}{2A}=\pm\frac{\sqrt{B^2-4AC}}{2A} ⇔ z = − B 2 A ± B 2 − 4 A C 2 A \Leftrightarrow z=-\frac{B}{2A}\pm\frac{\sqrt{B^2-4AC}}{2A} Next, notice the following B 2 − 4 A C = ( b 1 + i b 2 ) 2 − 4 ( a 1 + i a 2 ) ( c 1 + i c 2 ) B^2-4AC = (b_1+ib_2)^2-4(a_1+ia_2)(c_1+ic_2) = b 1 2 − b 2 2 + 2 i b 1 b 2 − 4 ( a 1 c 1 − a 2 c 2 ) − 4 i ( a 1 c 2 + a 2 c 1 ) =b_1^2-b_2^2+2ib_1b_2-4(a_1c_1-a_2c_2)-4i(a_1c_2+a_2c_1) = b 1 2 − b 2 2 − 4 ( a 1 c 1 − a 2 c 2 ) + 2 i ( b 1 b 2 − 2 ( a 1 c 2 + a 2 c 1 ) ) =b_1^2-b_2^2-4(a_1c_1-a_2c_2)+2i(b_1b_2-2(a_1c_2+a_2c_1)) = 0 + 2 i ( 0 ) = 0 , =0+2i(0) =0, since b 1 2 − b 2 2 = 4 ( a 1 c 1 − a 2 c 2 ) b_1^2-b_2^2 = 4(a_1c_1-a_2c_2) and b 1 b 2 = 2 ( a 1 c 2 + a 2 c 1 ) b_1b_2=2(a_1c_2+a_2c_1) . Therefore, x + i y = z = − B 2 A = − b 1 + i b 2 2 ( a 1 + i a 2 ) = − 1 2 ( a 1 − i a 2 ) ( b 1 + i b 2 ) ( a 1 − i a 2 ) ( a 1 + i a 2 ) x+iy=z=-\frac{B}{2A}=-\frac{b_1+ib_2}{2(a_1+ia_2)}=-\frac{1}{2}\frac{(a_1-ia_2)(b_1+ib_2)}{(a_1-ia_2)(a_1+ia_2)} = − 1 2 a 1 b 1 + a 2 b 2 + i ( a 1 b 2 − a 2 b 1 ) a 1 2 + a 2 2 = − 1 2 a 1 b 1 + a 2 b 2 a 1 2 + a 2 2 + i 1 2 a 2 b 1 − a 1 b 2 a 1 2 + a 2 2 =-\frac{1}{2}\frac{a_1b_1+a_2b_2+i(a_1b_2-a_2b_1)}{a_1^2+a_2^2}=-\frac{1}{2}\frac{a_1b_1+a_2b_2}{a_1^2+a_2^2}+i\frac{1}{2}\frac{a_2b_1-a_1b_2}{a_1^2+a_2^2} Since a 1 , a 2 , b 1 , b 2 , x , y a_1,a_2,b_1,b_2,x,y are all real, we can equate the real and imaginary parts of the equation. In particular, y = 1 2 a 2 b 1 − a 1 b 2 a 1 2 + a 2 2 y=\frac{1}{2}\frac{a_2b_1-a_1b_2}{a_1^2+a_2^2} So, m 2 − n 2 + p 2 − q 2 + r 2 − s 2 + t 2 − u 2 + v 2 = 2 2 − 2 2 + 1 2 − 1 2 + 2 2 − 1 2 + 2 2 − 2 2 + 2 2 = 7 m^2-n^2+p^2-q^2+r^2-s^2+t^2-u^2+v^2 = 2^2-2^2+1^2-1^2+2^2-1^2+2^2-2^2+2^2=7 .

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