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Algebra Difficulty 2.3 Junior Find the answer

Xiao Li uses the mathematical idea of substitution to solve the equation: (x2+1)2+4(x2+1)5=0(x^2+1)^2+4(x^2+1)-5=0. He treats (x2+1)(x^2+1) as a whole and sets x2+1=yx^2+1=y (y>0y>0), then the original equation can be transformed into y2+4y5=0y^2+4y-5=0, solving this yields y1=1y_1=1, y2=5y_2=-5 (which is discarded as it does not meet the condition). When y=1y=1, x2+1=1x^2+1=1, thus x2=0x^2=0, therefore x=0x=0. Hence, the solution to the original equation is x=0x=0. Please use this mathematical idea to solve the following problem:
In ABC\triangle ABC, C=90°\angle C=90°, the lengths of the two legs are aa and bb, and the length of the hypotenuse is cc, given that (a2+b2)(a2+b2+1)=12(a^2+b^2)(a^2+b^2+1)=12, find the length of the hypotenuse cc.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Let a2+b2=xa^2+b^2=x (x>0x>0), then (a2+b2)(a2+b2+1)=12(a^2+b^2)(a^2+b^2+1)=12 can be transformed into: x(x+1)=12x(x+1)=12, which simplifies to x2+x12=0x^2+x-12=0,
Solving this yields: x1=3x_1=3, x2=4x_2=-4 (which is discarded as it does not meet the condition),
Therefore, the value of a2+b2a^2+b^2 is 33,
Since C=90°\angle C=90°,
We have a2+b2=c2a^2+b^2=c^2,
Therefore, c2=3c^2=3,
Thus, c=3c=\sqrt{3}.

Answer: The length of the hypotenuse cc is 3\boxed{\sqrt{3}}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.