Is there any fraction crazier than this?

Algebra Level 3

Let a a , b b and c c be three distinct real constants. It is given that a 2 ( a − b ) ( a − c ) ( a + x ) + b 2 ( b − c ) ( b − a ) ( b + x ) + c 2 ( c − a ) ( c − b ) ( c + x ) ≡ p + q x + r x 2 ( a + x ) ( b + x ) ( c + x ) \frac { { a }^{ 2 } }{ \left( a-b \right) \left( a-c \right) \left( a+x \right) } +\frac { { b }^{ 2 } }{ \left( b-c \right) \left( b-a \right) \left( b+x \right) } +\frac { { c }^{ 2 } }{ \left( c-a \right) \left( c-b \right) \left( c+x \right) } \equiv \frac { p+qx+r{ x }^{ 2 } }{ \left( a+x \right) \left( b+x \right) \left( c+x \right) }

where p p , q q and r r are real constants. If s = 7 p + 8 q + 9 r s=7p+8q+9r , find the value of s


The answer is 9.

Raymond Chan by Raymond Chan , 20, USA

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1 solution

Raymond Chan
Nov 25, 2017

a 2 ( a − b ) ( a − c ) ( a + x ) + b 2 ( b − c ) ( b − a ) ( b + x ) + c 2 ( c − a ) ( c − b ) ( c + x ) ≡ p + q x + r x 2 ( a + x ) ( b + x ) ( c + x ) \frac { { a }^{ 2 } }{ \left( a-b \right) \left( a-c \right) \left( a+x \right) } +\frac { { b }^{ 2 } }{ \left( b-c \right) \left( b-a \right) \left( b+x \right) } +\frac { { c }^{ 2 } }{ \left( c-a \right) \left( c-b \right) \left( c+x \right) } \equiv \frac { p+qx+r{ x }^{ 2 } }{ \left( a+x \right) \left( b+x \right) \left( c+x \right) }

Multiply both sides by ( a + x ) ( b + x ) ( c + x ) \left( a+x \right) \left( b+x \right) \left( c+x \right) :

a 2 ( a + x ) ( b + x ) ( c + x ) ( a − b ) ( a − c ) ( a + x ) + b ( a + x ) ( b + x ) ( c + x ) ( b − c ) ( b − a ) ( b + x ) + c 2 ( a + x ) ( b + x ) ( c + x ) ( c − a ) ( c − b ) ( c + x ) ≡ p + q x + r x 2 \frac { { a }^{ 2 }\left( a+x \right) \left( b+x \right) \left( c+x \right) }{ \left( a-b \right) \left( a-c \right) \left( a+x \right) } +\frac { { b }\left( a+x \right) \left( b+x \right) \left( c+x \right) }{ \left( b-c \right) \left( b-a \right) \left( b+x \right) } +\frac { { c }^{ 2 }\left( a+x \right) \left( b+x \right) \left( c+x \right) }{ \left( c-a \right) \left( c-b \right) \left( c+x \right) } \equiv p+qx+r{ x }^{ 2 }

Simplify, we have:

a 2 ( b + x ) ( c + x ) ( a − b ) ( a − c ) + b ( a + x ) ( c + x ) ( b − c ) ( b − a ) + c 2 ( a + x ) ( b + x ) ( c − a ) ( c − b ) ≡ p + q x + r x 2 \frac { { a }^{ 2 }\left( b+x \right) \left( c+x \right) }{ \left( a-b \right) \left( a-c \right) } +\frac { { b }\left( a+x \right) \left( c+x \right) }{ \left( b-c \right) \left( b-a \right) } +\frac { { c }^{ 2 }\left( a+x \right) \left( b+x \right) }{ \left( c-a \right) \left( c-b \right) } \equiv p+qx+r{ x }^{ 2 }

Put x = − a x=-a : a 2 ( b − a ) ( c − a ) ( a − b ) ( a − c ) = p + q ( − a ) + r ( − a ) 2 \frac { { a }^{ 2 }\left( b-a \right) \left( c-a \right) }{ \left( a-b \right) \left( a-c \right) } =p+q\left( -a \right) +r{ \left( -a \right) }^{ 2 }

Simplify, we have:

r a 2 − q a + p = a 2 r{ a }^{ 2 }-qa+p={ a }^{ 2 }

Similarly, by putting x = − b x=-b and x = − c x=-c , we can obtain the following two equations: r b 2 − q b + p = b 2 r{ b }^{ 2 }-qb+p={ b }^{ 2 } r c 2 − q c + p = c 2 r{ c }^{ 2 }-qc+p={ c }^{ 2 }

Rearrange the three equations, we have: ( r − 1 ) a 2 − q a + p = 0 \left( r-1 \right) { a }^{ 2 }-qa+p=0 ( r − 1 ) b 2 − q b + p = 0 \left( r-1 \right) { b }^{ 2 }-qb+p=0 ( r − 1 ) c 2 − q c + p = 0 \left( r-1 \right) { c }^{ 2 }-qc+p=0

Note that the three equations we obtained share the same format ( r − 1 ) y 2 − q y + p = 0 \left( r-1 \right) { y }^{ 2 }-qy+p=0

Therefore, we can conclude that the equation ( r − 1 ) y 2 − q y + p = 0 \left( r-1 \right) { y }^{ 2 }-qy+p=0 has roots a a , b b and c c

However, a a , b b and c c are 3 distinct real constants, so it means the quadratic equation above has 3 distinct roots, which is impossible.

On the other hand, we know that the equation 0 y 2 − 0 y + 0 = 0 0{ y }^{ 2 }-0y+0=0 can has infinite many solutions.

Therefore, r − 1 = 0 r-1=0 , q = 0 q=0 and p = 0 p=0

Finally, we get r = 1 r=1 , q = 0 q=0 and p = 0 p=0

Thus, s = 7 p + 8 q + 9 r = 7 ( 0 ) + 8 ( 0 ) + 9 ( 1 ) = 9 s=7p+8q+9r=7(0)+8(0)+9(1)=\boxed9

1 Helpful 0 Interesting 0 Brilliant 0 Confused

A much faster way to solve this question is via cover up rule .

Pi Han Goh - 3 years, 6 months ago

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