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We Were Young Once ~ II  by Conquistadora 137 Review(s)
Starlight and MoonlightReviewed Chapter: 20 on 10/22/2018
It is wondrous the way you make your readers care for all the characters. It's not only with Thranduil and Oropher, but with all of your OCs. I keep worrying for Celebrin, I'm always afraid that Gwaelas will die, and with all of the other characters, too. I feel their joy and their sorrow as if it is my own, and that is a thing that is rarely found in not only fan fiction these days, but also in most other writing.

Starlight and MoonlightReviewed Chapter: 2 on 10/15/2018
I find it really great that you have a quote for every chapter. It must be kinda hard to find so many.

TariReviewed Chapter: 14 on 10/6/2016
Oropher is beginning to sound like Denethor (the movie version).

TariReviewed Chapter: 8 on 9/28/2016
I knew Annaroth was a spawn of Morgoth If not Morgoth himself. Brrrrrrrrr!

PSWReviewed Chapter: 26 on 1/1/2016
Very good, thanks for writing!

ZardiReviewed Chapter: 23 on 5/11/2014
I cannot even begin to describe to you how wonderful this story is and how much I am enjoying it. It is rare that I can read fanfiction and not think of one single criticism, yet here is one such story.

Your subtle handling of the first appearance of Annatar was masterful. I know who he is, but even if I had come into the story not knowing, those chapters still would have been successful. The suspense and looming evil were just perfect.

I have been waiting for Oropher to bite the dust. Not because I don't enjoy him as a character (I do...you've given him so much dimension), but because I am eager to see how you handle Thranduil the king. I only regret that you have not updated part three in years. I hope you won't abandon it forever.

Author Reply: Thanks for the lovely reviews! I promise I haven't abandoned part three; I've just hit a wall. I'm currently on a quest to simplify my life, so hopefully I'll have more time for writing in the future. :)

Reviewed Chapter: 26 on 1/19/2014
Wow...what can I say...wow...this part is so sad and dark! EXCELLENT writing! I have one question...what became of Thranduil's kinsman Amroth?

5yIvxpbUiReviewed Chapter: 12 on 10/23/2013
At last! Someone who unstdreands! Thanks for posting!

NoAjtqeZNReviewed Chapter: 10 on 10/23/2013
The plan is to have ready to serialize for late September of 2009.That would be ralley cool if you get to go to the JWHA conference!!! I expect that my brother will be there since he's very active in the group. Be sure to tell me if you do, and I'll tell him to look for you. And then blog about the whole thing -- I'll be happy to add you to ! :DI'll bet your kids are adorable! I have two myself (same two as back in my exmo-social days), and if you page thorough my blog, you can see how cute they are. ;^)

Q1P2Yw04kReviewed Chapter: 1 on 10/23/2013
Menon"3,4,5 എന ന വ വശങ ങള ആയ വര ന ന മട ട ത ര ക ണത ത ന റ Semi Perimeter 6 ആണല ല (3^3+4^3+5^3 = 6^3)6,8,10 എന ന വ വശങ ങള ആയ വര ന ന മട ട ത ര ക ണത ത ന റ Semi Perimeter 12 ആണല ല (6^3+8^3+10^3 = 12^39,12,15 എന ന വ വശങ ങള ആയ വര ന ന മട ട ത ര ക ണത ത ന റ Semi Perimeter 18 ആണല ല (9^3+12^3+15^3 = 18^3)ഈ ക ണ ന ന ബന ധത ത ന എന ത ങ ക ല ഒര ക രണ നല ക ക വ ന പറ റ മ എന ന ര സ ശയ ഉണ ട "May I approach the sauoititn as follows:3, 4 and 5 is a Pythagorean triplet.3^3 +4^3 + 5^3 = 27 + 64 + 125=216Also [(3+4+5)/2]^3 = 6^3 =216It proves that 3^3 +4^3 + 5^3 = [(3+4+5)/2]^3We notice that each successive Pythagorean triplets that you have mentioned in your post is an integral multiple of the triplet (3,4,5) and hence a general Pythagorean triplet in your post can be expressed as (3k, 4k, 5k)where k= 1, 2, 3, 4,.........Now (3k)^3 + (4k)^3 + (5k)^3 =k^3(3^3 +4^3+ 5^3)= k^3 [(3+4+5)/2]^3 by substitution from the previous relationHence, (3k)^3 + (4k)^3 + (5k)^3 =k^3 [(3+4+5)/2]^3 = [k(3+4+5)/2]^3 = [(3k+4k+5k)/2]^3i.e (3k)^3 + (4k)^3 + (5k)^3 = [(3k+4k+5k)/2]^3In the above relation put k=1, 2, 3, 4, 5..... Then we get all those relations mentioned in your post.This is a special property of Pythagorean triplets of the form (3k,4k,5k)If you examine the Pythagorean triplet (5,12,13)we see that this property namely sum of cubes equals cube of semi-perimeter does not hold.

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