Difference between revisions of "AoPS Wiki talk:Problem of the Day/July 17, 2011"
({{potd_solution}} is not needed after you write a solution!) |
|||
| Line 2: | Line 2: | ||
{{:AoPSWiki:Problem of the Day/July 17, 2011}} | {{:AoPSWiki:Problem of the Day/July 17, 2011}} | ||
==Solution== | ==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. | 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 10:48, 17 July 2011
Problem
AoPSWiki:Problem of the Day/July 17, 2011
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.
