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

({{potd_solution}} is not needed after you write a solution!) |
m (→Solution: link to Ptolemy's Theorem) |
||

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. |

## 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.