Crypto Art & Design Challenge [Round 9: IOTA]

sndbox (77)in #iota • 6 years ago (edited)

An Internet of All Things Art

Time to get your IOTA on! We've now hosted eight incredibly artful rounds of this crypto art challenge and received over 1,000+ jaw dropping artful entries from the Steem community. Last week was Stellar! (Definitely make sure to explore the results linked here and showcased below.) So, without further ado...

Welcome to the 9th Crypto Art and Design Challenge. Finally, this one will be visible - live through thecreativecrypto.com! We just launched The Creative Crypto Magazine (@creativecrypto) and with it, we're hosting some fun and creative crypto art exercises. To draw more into the conversation, we're building out a grassroots network via our Facebook Page and Twitter. Across all platforms - engagement towards crypto-based creativity has been amazing. It's been a lot of fun to build up an audience outside of Steemit and begin to reign in some crypto-curious internet-surfers back towards our hometown blockchain (Steem). With this ongoing series, we're calling all artists, graphic designers and craftspeople of the world.

In the spirit of crypto-creativity we ask that you -

Pick a famous artist (Picasso, Van Gogh, David Hockney, Michelangelo) whoever you want - and interpret the IOTA logo into that style. Other components of the image (background, details, other elements) are completely up to your imagination.

And now, IOTA!

Download and use the image below to get started:

Crypto Art Challenge Guidelines

Follow our magazine @creativecrypto! (Learn more, here.)
Create 1 image (1000 by 600px) using the IOTA logo (above)
Author a post explaining the Artist that you were inspired by
Publish a post using the hashtag #creativecrypto
Share a link to your post + image in the comments by Wednesday, 10pm EST

The top 5 creations will receive 20% votes from @sndbox and be showcased on The Creative Crypto media outlets. All entries that follow the guidelines will receive a partial vote (1% - 5%) depending on number of entries.