Go triangle!

"Take on the challenge of Go Triangle! Guide a speedy triangle as it soars forward, gathering all the dots in its path. Test your reflexes in this lively Android game. Tap the left or right side of the screen to steer the triangle in that direction. Avoid obstacles and collect dots to earn points and increase your score. Keep flying and scoring to beat your personal best. Hone your skills, grab bonuses, and aim for new records. Game highlights: Sleek, minimalist graphics Easy controls Top scores Engaging gameplay"<|endoftext|><|endoftext|><|endoftext|> # 2015 AMC 10A Problems/Problem 1. ## Contents. 1 Problem 2 Solution 3 Video Solution 4 See Also ## Problem What is the value of \[2-4+6-8+10-12+14-16+18-20?\] $\textbf{(A)}\ -10\qquad\textbf{(B)}\ -5\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 10$ ## Solution. We can rearrange the terms to get \[(2-4)+(6-8)+(10-12)+(14-16)+(18-20).\] Each of the terms in parentheses is equal to $-2$, so the expression simplifies to \[(-2)+(-2)+(-2)+(-2)+(-2)=-10.\] Therefore, the answer is $\boxed{\textbf{(A)}\ -10}$. ## Video Solution. https://youtu.be/8-3XeJXm-q0 ~savannahsolver <|endoftext|>## Mathematical Forums ## Category: High School Olympiads ## Topic: Inequality ## Views: 338 ## [enter: math-user1, num_posts=697, num_likes_received=372] ## [math-user1, num_likes=1] Let $a,b,c>0$. Prove that $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+3\sqrt[3]{abc}\geq 2(a+b+c)$ ## [enter: math-user2, num_posts=545, num_likes_received=72] ## [math-user2, num_likes=0] By AM-GM we have $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+3\sqrt[3]{abc}\ge4\sqrt[4]{\frac{a^2}{b}\frac{b^2}{c}\frac{c^2}{a}3\sqrt[3]{abc}}=4\sqrt[4]{3abc}$ and by AM-GM again we have $4\sqrt