Question Number 127788 by peter frank last updated on 02/Jan/21
Answered by mr W last updated on 02/Jan/21
![AB=[ab]=10a+b CD=[ba]=a+10b with 1≤a,b≤9 OE^2 =(1/4)(AB^2 −CD^2 ) ⇒OE=((√((AB+CD)(AB−CD)))/2) =((3(√(11(a+b)(a−b))))/2)=rational ⇒(a+b)(a−b)=11n^2 OE=((3×11n)/2)=((33n)/2) case 1: a+b=11 a−b=n^2 ⇒a=((11+n^2 )/2), b=((11−n^2 )/2) n=1: a=6, b=5 ⇒OE=((33)/2) n=3: a=10>9, b=1 ⇒bad case 2: a+b=n^2 a−b=11 ⇒a>11>9 ⇒bad case 3: a+b=11n a−b=n ⇒a=6n, b=5n n=1: a=6, b=5 ⇒OE=((33)/2) n=2: a=12>9, b=10>9 ⇒bad case 4: a+b=n a−b=11n>a+b ⇒bad case 5: a+b=11n^2 a−b=1 ⇒a=((11n^2 +1)/2), b=((11n^2 −1)/2) n=1: a=6, b=5 ⇒OE=((33)/2) ⇒the only solution is OE=((33)/2) with AB=65, CD=56](https://www.tinkutara.com/question/Q127794.png)
$${AB}=\left[{ab}\right]=\mathrm{10}{a}+{b} \\ $$$${CD}=\left[{ba}\right]={a}+\mathrm{10}{b} \\ $$$${with}\:\mathrm{1}\leqslant{a},{b}\leqslant\mathrm{9} \\ $$$$ \\ $$$${OE}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\left({AB}^{\mathrm{2}} −{CD}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{OE}=\frac{\sqrt{\left({AB}+{CD}\right)\left({AB}−{CD}\right)}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{3}\sqrt{\mathrm{11}\left({a}+{b}\right)\left({a}−{b}\right)}}{\mathrm{2}}={rational} \\ $$$$\Rightarrow\left({a}+{b}\right)\left({a}−{b}\right)=\mathrm{11}{n}^{\mathrm{2}} \\ $$$${OE}=\frac{\mathrm{3}×\mathrm{11}{n}}{\mathrm{2}}=\frac{\mathrm{33}{n}}{\mathrm{2}} \\ $$$$ \\ $$$${case}\:\mathrm{1}: \\ $$$${a}+{b}=\mathrm{11} \\ $$$${a}−{b}={n}^{\mathrm{2}} \\ $$$$\Rightarrow{a}=\frac{\mathrm{11}+{n}^{\mathrm{2}} }{\mathrm{2}},\:{b}=\frac{\mathrm{11}−{n}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${n}=\mathrm{1}:\:{a}=\mathrm{6},\:{b}=\mathrm{5}\:\Rightarrow{OE}=\frac{\mathrm{33}}{\mathrm{2}} \\ $$$${n}=\mathrm{3}:\:{a}=\mathrm{10}>\mathrm{9},\:{b}=\mathrm{1}\:\Rightarrow{bad} \\ $$$$ \\ $$$${case}\:\mathrm{2}: \\ $$$${a}+{b}={n}^{\mathrm{2}} \\ $$$${a}−{b}=\mathrm{11}\:\Rightarrow{a}>\mathrm{11}>\mathrm{9}\:\Rightarrow{bad} \\ $$$$ \\ $$$${case}\:\mathrm{3}: \\ $$$${a}+{b}=\mathrm{11}{n} \\ $$$${a}−{b}={n} \\ $$$$\Rightarrow{a}=\mathrm{6}{n},\:{b}=\mathrm{5}{n} \\ $$$${n}=\mathrm{1}:\:{a}=\mathrm{6},\:{b}=\mathrm{5}\:\Rightarrow{OE}=\frac{\mathrm{33}}{\mathrm{2}} \\ $$$${n}=\mathrm{2}:\:{a}=\mathrm{12}>\mathrm{9},\:{b}=\mathrm{10}>\mathrm{9}\:\Rightarrow{bad} \\ $$$$ \\ $$$${case}\:\mathrm{4}: \\ $$$${a}+{b}={n} \\ $$$${a}−{b}=\mathrm{11}{n}>{a}+{b}\:\Rightarrow{bad} \\ $$$$ \\ $$$${case}\:\mathrm{5}: \\ $$$${a}+{b}=\mathrm{11}{n}^{\mathrm{2}} \\ $$$${a}−{b}=\mathrm{1} \\ $$$$\Rightarrow{a}=\frac{\mathrm{11}{n}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}},\:{b}=\frac{\mathrm{11}{n}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}} \\ $$$${n}=\mathrm{1}:\:{a}=\mathrm{6},\:{b}=\mathrm{5}\:\Rightarrow{OE}=\frac{\mathrm{33}}{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow{the}\:{only}\:{solution}\:{is}\:{OE}=\frac{\mathrm{33}}{\mathrm{2}}\:{with} \\ $$$${AB}=\mathrm{65},\:{CD}=\mathrm{56} \\ $$
Commented by peter frank last updated on 02/Jan/21
$$\mathrm{thank}\:\mathrm{you}.\mathrm{happy}\:\mathrm{new}\:\mathrm{year} \\ $$
Commented by mr W last updated on 02/Jan/21
$${the}\:{same}\:{to}\:{you}! \\ $$