Difference between revisions of "AoPS Wiki talk:Problem of the Day/July 17, 2011"
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==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. |
Latest revision as of 18:33, 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.
