ok thanks
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May 30, 2020
0
#32
LLLhk wrote:
Day 5 problem coming:
Find the sum of the first 1000 cubes.
Sum of the first 1000 cubes = [1000(1000+1)/2]² ≈2.5050025x10¹¹
May 31, 2020
0
#34
From ARML 2020 Local that I took today:
TIEBREAKER PROBLEM
If triangle ABC has that AB=3, BC=5, and CA=7, and P is in the interior of triangle ABC such angle APB=angle BPC=angle CPA=120 degrees, then find PA+PB+PC
I thought ARML was cancelled? Unless ARML Local is different from ARML
i don't do questions everyday because now it is getting unactive but now that you are trying the problems I will try my best to do 1 question 1 day. (I try my best to do exactly when the new day starts at GMT time, as I wake up at 7:30 a.m. HKT and normally post at 8:00 a.m. HKT, which converts to exactly the start of a day in GMT time)