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Question Number 64469 by Tawa1 last updated on 18/Jul/19

	Answered by MJS last updated on 18/Jul/19
	$the radius of the incircle of a rectangular triangle is given by r=((a+b−(√(a^2 +b^2 )))/2) ((a+b−(√(a^2 +b^2 )))/2)=703 a+b−(√(a^2 +b^2 ))=1406 (√(a^2 +b^2 ))=a+b−1406 a^2 +b^2 =(a+b−1406)^2 2812a−2ab+2812b−1976836=0 ⇒ b=1406+((988418)/(a−1406)) put a=1406+t with t∈N^★ b=((988418)/t)+1406 988418=2×19^2 ×37^2 b∈N ⇒ t∣988418 we must try t∈{1, 2, 19, 37, 38, 74, 361, 703, 722, 1369, 1406, 2738, 13357, 26011, 26714, 52022, 494209, 988418} ⇒ with a<b<c [we can change a⇄b] a=1407 b=989724 c=989825 a=1408 b=495615 c=495617 a=1425 b=53428 c=53447 a=1443 b=28120 c=28156 a=1444 b=27417 c=27455 a=1480 b=14763 c=14837 a=1767 b=4144 c=4505 a=2109 b=2812 c=3515 a=2128 b=2775 c=3497$
	$the radius of the incircle of a rectangular$ $triangle is given by$ $r = \frac{a + b - \sqrt{a^{2} + b^{2}}}{2}$ $\frac{a + b - \sqrt{a^{2} + b^{2}}}{2} = 703$ $a + b - \sqrt{a^{2} + b^{2}} = 1406$ $\sqrt{a^{2} + b^{2}} = a + b - 1406$ $a^{2} + b^{2} = {(a + b - 1406)}^{2}$ $2812 a - 2 a b + 2812 b - 1976836 = 0$ $\Rightarrow b = 1406 + \frac{988418}{a - 1406}$ $put a = 1406 + t with t \in N^{★}$ $b = \frac{988418}{t} + 1406$ $988418 = 2 \times 19^{2} \times 37^{2}$ $b \in N \Rightarrow t ∣ 988418$ $we must try$ $t \in {1, 2, 19, 37, 38, 74, 361, 703, 722, 1369, 1406, 2738, 13357, 26011, 26714, 52022, 494209, 988418}$ $\Rightarrow$ $with a < b < c [we can change a ⇄ b]$ $a = 1407 b = 989724 c = 989825$ $a = 1408 b = 495615 c = 495617$ $a = 1425 b = 53428 c = 53447$ $a = 1443 b = 28120 c = 28156$ $a = 1444 b = 27417 c = 27455$ $a = 1480 b = 14763 c = 14837$ $a = 1767 b = 4144 c = 4505$ $a = 2109 b = 2812 c = 3515$ $a = 2128 b = 2775 c = 3497$
	Commented by Prithwish sen last updated on 18/Jul/19

	$great!$
	Commented by MJS last updated on 18/Jul/19

	$I will post {Heron}^{'} s formula and its conclusions$ $as soon as I can find the time$
	Commented by Prithwish sen last updated on 18/Jul/19

	$Sir how could one be good at geometry . What$ $is the path one should follow ?$
	Commented by Prithwish sen last updated on 18/Jul/19

	$MJS Sir please {don}^{'} t ignore . Suggest something .$
	Commented by Tawa1 last updated on 18/Jul/19

	$God bless you sirs, i appreciate your effort$
	Commented by mr W last updated on 18/Jul/19

	$A r e a o f r e c t a n g u l a r t r i a n g l e A$ $A = \frac{a b}{2}$ $A = \frac{r (a + b + c)}{2} = \frac{r (a + b + \sqrt{a^{2} + b^{2}})}{2}$ $\frac{r (a + b + \sqrt{a^{2} + b^{2}})}{2} = \frac{a b}{2}$ $\Rightarrow r = \frac{a b}{a + b + \sqrt{a^{2} + b^{2}}}$ $\Rightarrow r = \frac{a b (a + b - \sqrt{a^{2} + b^{2}})}{(a + b + \sqrt{a^{2} + b^{2}}) (a + b - \sqrt{a^{2} + b^{2}})}$ $\Rightarrow r = \frac{a b (a + b - \sqrt{a^{2} + b^{2}})}{2 a b}$ $\Rightarrow r = \frac{a + b - \sqrt{a^{2} + b^{2}}}{2}$
	Commented by MJS last updated on 18/Jul/19

	$sorry no time at the moment$
	Commented by Prithwish sen last updated on 18/Jul/19

	$ok thank you sir .$
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