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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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