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Question Number 170212 by otchereabdullai@gmail.com last updated on 18/May/22

 which rectangle with integer length     side have numerically the same area    and perimeter? Find them all. Find   a proof that convinces that you have   found them all. what about right−   angled triangle? how many solutions?

$$\:\mathrm{which}\:\mathrm{rectangle}\:\mathrm{with}\:\mathrm{integer}\:\mathrm{length}\: \\ $$$$\:\:\mathrm{side}\:\mathrm{have}\:\mathrm{numerically}\:\mathrm{the}\:\mathrm{same}\:\mathrm{area} \\ $$$$\:\:\mathrm{and}\:\mathrm{perimeter}?\:\mathrm{Find}\:\mathrm{them}\:\mathrm{all}.\:\mathrm{Find} \\ $$$$\:\mathrm{a}\:\mathrm{proof}\:\mathrm{that}\:\mathrm{convinces}\:\mathrm{that}\:\mathrm{you}\:\mathrm{have} \\ $$$$\:\mathrm{found}\:\mathrm{them}\:\mathrm{all}.\:\mathrm{what}\:\mathrm{about}\:\mathrm{right}− \\ $$$$\:\mathrm{angled}\:\mathrm{triangle}?\:\mathrm{how}\:\mathrm{many}\:\mathrm{solutions}? \\ $$

Answered by aleks041103 last updated on 18/May/22

rectangle − a,b  ⇒ab=a+b  ab−a−b+1=1  a(b−1)−(b−1)=1  (a−1)(b−1)=1  a,b≥0 and are integers  ⇒a−1∣1 and b−1∣1  but of all natural numbers, only 1∣1.  ⇒a−1=b−1=1⇒a=b=2  ⇒Square of sidelength 2 is the only  soln.

$${rectangle}\:−\:{a},{b} \\ $$$$\Rightarrow{ab}={a}+{b} \\ $$$${ab}−{a}−{b}+\mathrm{1}=\mathrm{1} \\ $$$${a}\left({b}−\mathrm{1}\right)−\left({b}−\mathrm{1}\right)=\mathrm{1} \\ $$$$\left({a}−\mathrm{1}\right)\left({b}−\mathrm{1}\right)=\mathrm{1} \\ $$$${a},{b}\geqslant\mathrm{0}\:{and}\:{are}\:{integers} \\ $$$$\Rightarrow{a}−\mathrm{1}\mid\mathrm{1}\:{and}\:{b}−\mathrm{1}\mid\mathrm{1} \\ $$$${but}\:{of}\:{all}\:{natural}\:{numbers},\:{only}\:\mathrm{1}\mid\mathrm{1}. \\ $$$$\Rightarrow{a}−\mathrm{1}={b}−\mathrm{1}=\mathrm{1}\Rightarrow{a}={b}=\mathrm{2} \\ $$$$\Rightarrow{Square}\:{of}\:{sidelength}\:\mathrm{2}\:{is}\:{the}\:{only} \\ $$$${soln}. \\ $$$$\: \\ $$

Commented by Rasheed.Sindhi last updated on 18/May/22

∩i⊂∈!

$$\cap\mathrm{i}\subset\in! \\ $$

Commented by otchereabdullai@gmail.com last updated on 18/May/22

God bless you sir am much grateful  for your time!

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{am}\:\mathrm{much}\:\mathrm{grateful} \\ $$$$\mathrm{for}\:\mathrm{your}\:\mathrm{time}! \\ $$

Commented by floor(10²Eta[1]) last updated on 18/May/22

the perimeter is 2a+2b

$$\mathrm{the}\:\mathrm{perimeter}\:\mathrm{is}\:\mathrm{2a}+\mathrm{2b} \\ $$$$ \\ $$

Commented by floor(10²Eta[1]) last updated on 18/May/22

man if the side of the square is 2 the perimeter is 8

$$\mathrm{man}\:\mathrm{if}\:\mathrm{the}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\:\mathrm{is}\:\mathrm{2}\:\mathrm{the}\:\mathrm{perimeter}\:\mathrm{is}\:\mathrm{8} \\ $$

Commented by aleks041103 last updated on 19/May/22

Yes, you′re right. Sorry!

$${Yes},\:{you}'{re}\:{right}.\:{Sorry}! \\ $$

Answered by aleks041103 last updated on 18/May/22

for rigth triangles:  want: xy=2(x+y+(√(x^2 +y^2 )))  since x,y are integers, we need (√(x^2 +y^2 )) to  be an integer. This can happen if we   have a pythagorian triple.  All of them are generated by:  x=u^2 −w^2   y=2uw  (√(x^2 +y^2 ))=u^2 +w^2   ⇒(u^2 −w^2 )2uw=2(u^2 −w^2 +2uw+u^2 +w^2 )  ⇒uw(u^2 −w^2 )=2u(u+w)  ⇒w(u−w)(u+w)=2(u+w)  ⇒w(u−w)=2  Since x,y∈N, u,w∈N  ⇒w∣2 and u−w∣2  ⇒w=1,2 and u−w=2,1  ⇒(u,w)∈{(3,1),(3,2)}  ⇒(x,y)∈{(8,6),(5,12)}  and really  •x=8,y=6,z=(√(x^2 +y^2 ))=10  (1/2)xy=24=6+8+10✓  •x=5,y=12,z=13  (1/2)xy=30=5+12+13✓

$${for}\:{rigth}\:{triangles}: \\ $$$${want}:\:{xy}=\mathrm{2}\left({x}+{y}+\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right) \\ $$$${since}\:{x},{y}\:{are}\:{integers},\:{we}\:{need}\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:{to} \\ $$$${be}\:{an}\:{integer}.\:{This}\:{can}\:{happen}\:{if}\:{we}\: \\ $$$${have}\:{a}\:{pythagorian}\:{triple}. \\ $$$${All}\:{of}\:{them}\:{are}\:{generated}\:{by}: \\ $$$${x}={u}^{\mathrm{2}} −{w}^{\mathrm{2}} \\ $$$${y}=\mathrm{2}{uw} \\ $$$$\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }={u}^{\mathrm{2}} +{w}^{\mathrm{2}} \\ $$$$\Rightarrow\left({u}^{\mathrm{2}} −{w}^{\mathrm{2}} \right)\mathrm{2}{uw}=\mathrm{2}\left({u}^{\mathrm{2}} −{w}^{\mathrm{2}} +\mathrm{2}{uw}+{u}^{\mathrm{2}} +{w}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{uw}\left({u}^{\mathrm{2}} −{w}^{\mathrm{2}} \right)=\mathrm{2}{u}\left({u}+{w}\right) \\ $$$$\Rightarrow{w}\left({u}−{w}\right)\left({u}+{w}\right)=\mathrm{2}\left({u}+{w}\right) \\ $$$$\Rightarrow{w}\left({u}−{w}\right)=\mathrm{2} \\ $$$${Since}\:{x},{y}\in\mathbb{N},\:{u},{w}\in\mathbb{N} \\ $$$$\Rightarrow{w}\mid\mathrm{2}\:{and}\:{u}−{w}\mid\mathrm{2} \\ $$$$\Rightarrow{w}=\mathrm{1},\mathrm{2}\:{and}\:{u}−{w}=\mathrm{2},\mathrm{1} \\ $$$$\Rightarrow\left({u},{w}\right)\in\left\{\left(\mathrm{3},\mathrm{1}\right),\left(\mathrm{3},\mathrm{2}\right)\right\} \\ $$$$\Rightarrow\left({x},{y}\right)\in\left\{\left(\mathrm{8},\mathrm{6}\right),\left(\mathrm{5},\mathrm{12}\right)\right\} \\ $$$${and}\:{really} \\ $$$$\bullet{x}=\mathrm{8},{y}=\mathrm{6},{z}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }=\mathrm{10} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{xy}=\mathrm{24}=\mathrm{6}+\mathrm{8}+\mathrm{10}\checkmark \\ $$$$\bullet{x}=\mathrm{5},{y}=\mathrm{12},{z}=\mathrm{13} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{xy}=\mathrm{30}=\mathrm{5}+\mathrm{12}+\mathrm{13}\checkmark \\ $$

Commented by Rasheed.Sindhi last updated on 18/May/22

Nice!

$$\mathcal{N}{ice}! \\ $$

Commented by otchereabdullai@gmail.com last updated on 18/May/22

God bless you sir am much grateful    for your time!

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{am}\:\mathrm{much}\:\mathrm{grateful}\: \\ $$$$\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}! \\ $$

Answered by floor(10²Eta[1]) last updated on 18/May/22

ab=2a+2b  2a−ab+2b=0  a(2−b)−2(2−b)=−4  (2−a)(2−b)=4  2−a=−1  2−b=−4  a=3, b=6  2−a=−2  2−b=−2  a=b=4  all sol: {(4,4),(3,6),(6,3)}

$$\mathrm{ab}=\mathrm{2a}+\mathrm{2b} \\ $$$$\mathrm{2a}−\mathrm{ab}+\mathrm{2b}=\mathrm{0} \\ $$$$\mathrm{a}\left(\mathrm{2}−\mathrm{b}\right)−\mathrm{2}\left(\mathrm{2}−\mathrm{b}\right)=−\mathrm{4} \\ $$$$\left(\mathrm{2}−\mathrm{a}\right)\left(\mathrm{2}−\mathrm{b}\right)=\mathrm{4} \\ $$$$\mathrm{2}−\mathrm{a}=−\mathrm{1} \\ $$$$\mathrm{2}−\mathrm{b}=−\mathrm{4} \\ $$$$\mathrm{a}=\mathrm{3},\:\mathrm{b}=\mathrm{6} \\ $$$$\mathrm{2}−\mathrm{a}=−\mathrm{2} \\ $$$$\mathrm{2}−\mathrm{b}=−\mathrm{2} \\ $$$$\mathrm{a}=\mathrm{b}=\mathrm{4} \\ $$$$\mathrm{all}\:\mathrm{sol}:\:\left\{\left(\mathrm{4},\mathrm{4}\right),\left(\mathrm{3},\mathrm{6}\right),\left(\mathrm{6},\mathrm{3}\right)\right\} \\ $$

Commented by Rasheed.Sindhi last updated on 18/May/22

∨. ∩i⊂∈!

$$\vee.\:\cap\mathrm{i}\subset\in! \\ $$

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