Question-201150

Question Number 201150 by Mingma last updated on 30/Nov/23

Answered by mr W last updated on 02/Dec/23

a=side length of square (((a^2 +x^2 −15^2 )/(2ax)))^2 +(((a^2 +x^2 −20^2 )/(2ax)))^2 =1 a^4 +x^4 −(15^2 +20^2 )(a^2 +x^2 )+((15^4 +20^4 )/2)=0 x^4 −(15^2 +20^2 )x^2 +a^4 −(15^2 +20^2 )a^2 +((15^4 +20^4 )/2)=0 (((15^2 +20^2 )^2 )/4)≥x^4 −(15^2 +20^2 )x^2 +((15^4 +20^4 )/2) x^4 −(15^2 +20^2 )x^2 +(((20^2 −15^2 )^2 )/4)≤0 (((20−15)^2 )/2)≤x^2 ≤(((20+15)^2 )/2) ⇒((20−15)/( (√2)))≤x≤((20+15)/( (√2))) x^2 +y^2 =15^2 +20^2 =625 for x, y∈N there are 2 possibilities: (x, y)=(15, 20) or (7, 24) a^2 =(1/2)[625±(√(4x^2 (625−x^2 )−30625))] ⇒a^2 =((625±25(√(527)))/2) or ((625±7(√(1679)))/2) if we only consider the cases that the point lies inside the square, then there are two solutions: (x, y)=(15, 20) and the area of the square is ((625+25(√(527)))/2) or (x, y)=(7, 24) and the area of the square is ((625+7(√(1679)))/2).

$${a}={side}\:{length}\:{of}\:{square} \\ $$$$\left(\frac{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{15}^{\mathrm{2}} }{\mathrm{2}{ax}}\right)^{\mathrm{2}} +\left(\frac{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{20}^{\mathrm{2}} }{\mathrm{2}{ax}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${a}^{\mathrm{4}} +{x}^{\mathrm{4}} −\left(\mathrm{15}^{\mathrm{2}} +\mathrm{20}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)+\frac{\mathrm{15}^{\mathrm{4}} +\mathrm{20}^{\mathrm{4}} }{\mathrm{2}}=\mathrm{0} \\ $$$${x}^{\mathrm{4}} −\left(\mathrm{15}^{\mathrm{2}} +\mathrm{20}^{\mathrm{2}} \right){x}^{\mathrm{2}} +{a}^{\mathrm{4}} −\left(\mathrm{15}^{\mathrm{2}} +\mathrm{20}^{\mathrm{2}} \right){a}^{\mathrm{2}} +\frac{\mathrm{15}^{\mathrm{4}} +\mathrm{20}^{\mathrm{4}} }{\mathrm{2}}=\mathrm{0} \\ $$$$\frac{\left(\mathrm{15}^{\mathrm{2}} +\mathrm{20}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}}\geqslant{x}^{\mathrm{4}} −\left(\mathrm{15}^{\mathrm{2}} +\mathrm{20}^{\mathrm{2}} \right){x}^{\mathrm{2}} +\frac{\mathrm{15}^{\mathrm{4}} +\mathrm{20}^{\mathrm{4}} }{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} −\left(\mathrm{15}^{\mathrm{2}} +\mathrm{20}^{\mathrm{2}} \right){x}^{\mathrm{2}} +\frac{\left(\mathrm{20}^{\mathrm{2}} −\mathrm{15}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}}\leqslant\mathrm{0} \\ $$$$\frac{\left(\mathrm{20}−\mathrm{15}\right)^{\mathrm{2}} }{\mathrm{2}}\leqslant{x}^{\mathrm{2}} \leqslant\frac{\left(\mathrm{20}+\mathrm{15}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{20}−\mathrm{15}}{\:\sqrt{\mathrm{2}}}\leqslant{x}\leqslant\frac{\mathrm{20}+\mathrm{15}}{\:\sqrt{\mathrm{2}}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{15}^{\mathrm{2}} +\mathrm{20}^{\mathrm{2}} =\mathrm{625} \\ $$$${for}\:{x},\:{y}\in{N}\:{there}\:{are}\:\mathrm{2}\:{possibilities}: \\ $$$$\left({x},\:{y}\right)=\left(\mathrm{15},\:\mathrm{20}\right)\:{or}\:\left(\mathrm{7},\:\mathrm{24}\right) \\ $$$${a}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{625}\pm\sqrt{\mathrm{4}{x}^{\mathrm{2}} \left(\mathrm{625}−{x}^{\mathrm{2}} \right)−\mathrm{30625}}\right] \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\frac{\mathrm{625}\pm\mathrm{25}\sqrt{\mathrm{527}}}{\mathrm{2}}\:{or}\:\frac{\mathrm{625}\pm\mathrm{7}\sqrt{\mathrm{1679}}}{\mathrm{2}} \\ $$$${if}\:{we}\:{only}\:{consider}\:{the}\:{cases}\:{that} \\ $$$${the}\:{point}\:{lies}\:{inside}\:{the}\:{square},\:{then} \\ $$$${there}\:{are}\:{two}\:{solutions}: \\ $$$$\left({x},\:{y}\right)=\left(\mathrm{15},\:\mathrm{20}\right)\:{and}\:{the}\:{area}\:{of}\:{the} \\ $$$${square}\:{is}\:\frac{\mathrm{625}+\mathrm{25}\sqrt{\mathrm{527}}}{\mathrm{2}}\:{or} \\ $$$$\left({x},\:{y}\right)=\left(\mathrm{7},\:\mathrm{24}\right)\:{and}\:{the}\:{area}\:{of}\:{the} \\ $$$${square}\:{is}\:\frac{\mathrm{625}+\mathrm{7}\sqrt{\mathrm{1679}}}{\mathrm{2}}. \\ $$

Commented by mr W last updated on 02/Dec/23