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Question Number 158619 by amin96 last updated on 06/Nov/21

Commented by amin96 last updated on 06/Nov/21


	Answered by Rasheed.Sindhi last updated on 07/Nov/21
	$AX=BY=CZ=h (Assuming h integer) •▲_(XYZ) =(√(S(S−XY)(S−YZ)(S−ZX))) XZ=(√((12−h)^2 +h^2 )) =(√(144−24h+2h^2 )) XY=(√((9−h)^2 +h^2 −2(9−h)(h)cosB)) cosB=(9/(15)) =(3/5) XY=(√((9−h)^2 +h^2 −2(9−h)(h)((3/5)))) =(√(81−18h+h^2 +h^2 −18.(3/5)h+2.(3/5)h^2 )) =(√(81−18h+2h^2 −((54)/5)h+(6/5)h^2 )) =(√((405−90h+10h^2 −54h+6h^2 )/5)) =(√((16h^2 −144h+405)/5)) { ((XZ=(√(2(72−12h+h^2 ))))),((XY=(√((16h^2 −144h+405)/5)))) :} For being 2(72−12h+h^2 ) perfect square, 72−12h+h^2 ∈E⇒h∈E For being ((16h^2 −144h+405)/5) perfect square, 5∣h ∵ 2∣h ∧ 5∣h⇒10∣h ∴ h has no integral value within range$
	$A X = B Y = C Z = h (A s s u m i n g h i n t e g e r)$ $∙ ▴_{X Y Z} = \sqrt{S (S - X Y) (S - Y Z) (S - Z X)}$ $X Z = \sqrt{{(12 - h)}^{2} + h^{2}}$ $= \sqrt{144 - 24 h + 2 h^{2}}$ $X Y = \sqrt{{(9 - h)}^{2} + h^{2} - 2 (9 - h) (h) \cos B}$ $\cos B = \frac{9}{15} = \frac{3}{5}$ $X Y = \sqrt{{(9 - h)}^{2} + h^{2} - 2 (9 - h) (h) (\frac{3}{5})}$ $= \sqrt{81 - 18 h + h^{2} + h^{2} - 18 . \frac{3}{5} h + 2 . \frac{3}{5} h^{2}}$ $= \sqrt{81 - 18 h + 2 h^{2} - \frac{54}{5} h + \frac{6}{5} h^{2}}$ $= \sqrt{\frac{405 - 90 h + 10 h^{2} - 54 h + 6 h^{2}}{5}}$ $= \sqrt{\frac{16 h^{2} - 144 h + 405}{5}}$ ${\begin{cases} X Z = \sqrt{2 (72 - 12 h + h^{2})} \\ X Y = \sqrt{\frac{16 h^{2} - 144 h + 405}{5}} \end{cases}$ $F o r b e i n g 2 (72 - 12 h + h^{2}) p e r f e c t$ $s q u a r e, 72 - 12 h + h^{2} \in E \Rightarrow h \in E$ $F o r b e i n g \frac{16 h^{2} - 144 h + 405}{5} p e r f e c t$ $s q u a r e, 5 ∣ h$ $∵ 2 ∣ h \land 5 ∣ h \Rightarrow 10 ∣ h$ $∴ h h a s n o i n t e g r a l v a l u e w i t h i n r a n g e$
	Commented by amin96 last updated on 07/Nov/21

	Is XYZ a right triangle?
	Commented by Rasheed.Sindhi last updated on 07/Nov/21

	$S o r r y I^{'} l l e d i t m y a n s w e r, b u t t h e a n s w e r,$ $I t h i n k, w i l l n o t c h a n g e . O n l y t h e$ $a r e a f o r m u l a w i l l c h a n g e .$
	Commented by amin96 last updated on 07/Nov/21

	Yes, I think so too. So there is no integer for the area of the yellow triangle?
	Commented by mr W last updated on 07/Nov/21

	$o n l y t h e v a l u e o f t h e a r e a o f t h e$ $y e l l o w t r i a n g l e s h o u l d b e i n t e g e r,$ $n o t t h e s i d e l e n g t h o f t h e t r i a n g l e s!$
	Answered by mr W last updated on 07/Nov/21

	$A B = p = 9, A C = q = 12, B C = r = 15$ $l e t A X = B Y = C Z = t$ $l e t Δ = a r e a o f Δ A B C = 54$ $Δ_{A Z X} = \frac{t}{p} \times \frac{(q - t)}{q} \times Δ$ $Δ_{B X Y} = \frac{t}{r} \times \frac{(p - t)}{p} \times Δ$ $Δ_{C Y Z} = \frac{t}{q} \times \frac{(r - t)}{r} \times Δ$ $Δ_{X Y Z} = [1 - \frac{t (q - t)}{p q} - \frac{t (p - t)}{r p} - \frac{t (r - t)}{q r}] Δ$ $Δ_{X Y Z} = \frac{(p + q + r) t^{2} - (p q + q r + r p) t + p q r}{p q r} Δ$ $Δ_{X Y Z} = \frac{3 (4 t^{2} - 47 t + 180)}{10}$ $Δ_{X Y Z} = \frac{3 t (4 t - 47)}{10} + 54$ $\frac{3 t (4 t - 47)}{10} m u s t b e a n i n t e g e r k .$ $Δ_{X Y Z} = k + 54$ $12 t^{2} - 141 t = 10 k w i t h k \in Z$ $t = \frac{141 \pm \sqrt{141^{2} + 480 k}}{24}$ $141^{2} + 480 k ⩾ 0 \Rightarrow k ⩾ - \frac{141^{2}}{480} \Rightarrow k ⩾ - 41$ $s u c h t h a t 0 < t < 9 \Rightarrow - 41 ⩽ k ⩽ - 1$ $t h e r e f o r e w e h a v e f o l l o w i n g$ $i n t e g e r a r e a f o r Δ X Y Z : k + 54$ $- 41 ⩽ k ⩽ - 30 f o r e a c h k t w o t r i a n g l e s e x i s t .$ $- 29 ⩽ k ⩽ - 1 f o r e a c h k o n e t r i a n g l e e x i s t s .$ $t h a t m e a n s t o t a l l y 29 + 2 \times 12 = 43$ $p o s s i b l e y e l l o w t r i a n g l e s e x i s t .$ $s u m o f a l l p o s s i b l e i n t e g e r a r e a s :$ $41 \times 54 + (- 41 - 40 - 39 - . . . - 1)$ $= 41 \times 54 - \frac{42 \times 41}{2} = 1335$
	Commented by mr W last updated on 07/Nov/21

	$a n o t h e r w a y :$ $m a x . y e l l o w t r i a n g l e s h o u l d b e$ $s m a l l e r t h a n Δ A B C = 54$ $m i n . y e l l o w t r i a n g l e (s e e Q 158664)$ $= (1 - \frac{{(9 \times 12 + 12 \times 15 + 15 \times 9)}^{2}}{4 \times 9 \times 12 \times 15 \times (9 + 12 + 15)}) \times 54$ $\approx 12.6$ $t h a t m e a n s t h e i n t e g e r v a l u e f o r$ $t h e a r e a o f t h e y e l l o w t r i a n g l e c a n$ $b e f r o m 13 t o 53 .$ $s u m o f t h e s e v a l u e s$ $= \frac{(13 + 53) \times 41}{2} = 1353$
	Commented by Tawa11 last updated on 07/Nov/21

	$Great sir$
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