Difference between revisions of "AoPS Wiki talk:Problem of the Day/July 17, 2011"
(new problem talk page) |
|||
Line 3: | Line 3: | ||
==Solution== | ==Solution== | ||
{{potd_solution}} | {{potd_solution}} | ||
+ | By the Ptolemy's Theorem, we may say (MN x OP) + (NO x PM) = MO x NP. Substituting the values given, (3 x 5) + (4 x 3) = 6 x NP. 15 + 12 = 27 = 6NP. Thus, NP = 27/6 = 9/2. |
Revision as of 09:27, 17 July 2011
Problem
AoPSWiki:Problem of the Day/July 17, 2011
Solution
This Problem of the Day needs a solution. If you have a solution for it, please help us out by adding it. By the Ptolemy's Theorem, we may say (MN x OP) + (NO x PM) = MO x NP. Substituting the values given, (3 x 5) + (4 x 3) = 6 x NP. 15 + 12 = 27 = 6NP. Thus, NP = 27/6 = 9/2.