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?


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

$which rectangle with integer length$ $side have numerically the same area$ $and perimeter ? Find them all . Find$ $a proof that convinces that you have$ $found them all . what about right -$ $angled triangle ? how many solutions ?$
	Answered by aleks041103 last updated on 18/May/22

	$r e c t a n g l e - a, b$ $\Rightarrow a b = a + b$ $a b - a - b + 1 = 1$ $a (b - 1) - (b - 1) = 1$ $(a - 1) (b - 1) = 1$ $a, b ⩾ 0 a n d a r e i n t e g e r s$ $\Rightarrow a - 1 ∣ 1 a n d b - 1 ∣ 1$ $b u t o f a l l n a t u r a l n u m b e r s, o n l y 1 ∣ 1 .$ $\Rightarrow a - 1 = b - 1 = 1 \Rightarrow a = b = 2$ $\Rightarrow S q u a r e o f s i d e l e n g t h 2 i s t h e o n l y$ $s o l n .$
	Commented by Rasheed.Sindhi last updated on 18/May/22

	$\cap i \subset\in!$
	Commented by otchereabdullai@gmail.com last updated on 18/May/22

	$God bless you sir am much grateful$ $for your time!$
	Commented by floor(10²Eta[1]) last updated on 18/May/22

	$the perimeter is 2 a + 2 b$
	Commented by floor(10²Eta[1]) last updated on 18/May/22

	$man if the side of the square is 2 the perimeter is 8$
	Commented by aleks041103 last updated on 19/May/22

	$Y e s, {y o u}^{'} r e r i g h t . S o r r y!$
	Answered by aleks041103 last updated on 18/May/22
	$for rigth triangles: want: xy=2(x+y+(√(x^2 +y^2 ))) since x,y are integers, we need (√(x^2 +y^2 )) to be an integer. This can happen if we have a pythagorian triple. All of them are generated by: x=u^2 −w^2 y=2uw (√(x^2 +y^2 ))=u^2 +w^2 ⇒(u^2 −w^2 )2uw=2(u^2 −w^2 +2uw+u^2 +w^2 ) ⇒uw(u^2 −w^2 )=2u(u+w) ⇒w(u−w)(u+w)=2(u+w) ⇒w(u−w)=2 Since x,y∈N, u,w∈N ⇒w∣2 and u−w∣2 ⇒w=1,2 and u−w=2,1 ⇒(u,w)∈{(3,1),(3,2)} ⇒(x,y)∈{(8,6),(5,12)} and really •x=8,y=6,z=(√(x^2 +y^2 ))=10 (1/2)xy=24=6+8+10✓ •x=5,y=12,z=13 (1/2)xy=30=5+12+13✓$
	$f o r r i g t h t r i a n g l e s :$ $w a n t : x y = 2 (x + y + \sqrt{x^{2} + y^{2}})$ $s i n c e x, y a r e i n t e g e r s, w e n e e d \sqrt{x^{2} + y^{2}} t o$ $b e a n i n t e g e r . T h i s c a n h a p p e n i f w e$ $h a v e a p y t h a g o r i a n t r i p l e .$ $A l l o f t h e m a r e g e n e r a t e d b y :$ $x = u^{2} - w^{2}$ $y = 2 u w$ $\sqrt{x^{2} + y^{2}} = u^{2} + w^{2}$ $\Rightarrow (u^{2} - w^{2}) 2 u w = 2 (u^{2} - w^{2} + 2 u w + u^{2} + w^{2})$ $\Rightarrow u w (u^{2} - w^{2}) = 2 u (u + w)$ $\Rightarrow w (u - w) (u + w) = 2 (u + w)$ $\Rightarrow w (u - w) = 2$ $S i n c e x, y \in N, u, w \in N$ $\Rightarrow w ∣ 2 a n d u - w ∣ 2$ $\Rightarrow w = 1, 2 a n d u - w = 2, 1$ $\Rightarrow (u, w) \in {(3, 1), (3, 2)}$ $\Rightarrow (x, y) \in {(8, 6), (5, 12)}$ $a n d r e a l l y$ $∙ x = 8, y = 6, z = \sqrt{x^{2} + y^{2}} = 10$ $\frac{1}{2} x y = 24 = 6 + 8 + 10 ✓$ $∙ x = 5, y = 12, z = 13$ $\frac{1}{2} x y = 30 = 5 + 12 + 13 ✓$
	Commented by Rasheed.Sindhi last updated on 18/May/22

	$N i c e!$
	Commented by otchereabdullai@gmail.com last updated on 18/May/22

	$God bless you sir am much grateful$ $for your time!$
	Answered by floor(10²Eta[1]) last updated on 18/May/22

	$ab = 2 a + 2 b$ $2 a - ab + 2 b = 0$ $a (2 - b) - 2 (2 - b) = - 4$ $(2 - a) (2 - b) = 4$ $2 - a = - 1$ $2 - b = - 4$ $a = 3, b = 6$ $2 - a = - 2$ $2 - b = - 2$ $a = b = 4$ $all sol : {(4, 4), (3, 6), (6, 3)}$
	Commented by Rasheed.Sindhi last updated on 18/May/22

	$\lor . \cap i \subset\in!$
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