Difference between revisions of "AoPS Wiki talk:Problem of the Day/July 17, 2011"

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{{: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.
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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.

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