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Triangle $ABC$ has altitudes $\overline{AD},$ $\overline{BE},$ and $\overline{CF}.$  If $AD = 12,$ $BE = 20,$ and $CF$ is a positive integer, then find the largest possible value of $CF.$

 Dec 10, 2023
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Since a triangle's area is given by 21​⋅base⋅height, we can express the areas of triangles ABD, BCE, and CAF as 21​⋅12⋅AD, 21​⋅20⋅BE, and 21​⋅CF⋅CF, respectively.

 

Since triangles ABD, BCE, and CAF comprise all of triangle ABC, their areas must sum to the area of triangle ABC. Thus, we have 21​⋅12⋅AD+21​⋅20⋅BE+21​⋅CF2=21​⋅AB⋅BC.

 

Substituting the given values of AD and BE, we get 6⋅12+10⋅20+21​CF2=21​AB⋅BC.

 

This simplifies to 72+200+21​CF2=21​AB⋅BC. Rearranging and multiplying by 2, we get 344+CF2=AB⋅BC.

 

Since CF is a positive integer, CF must be less than or equal to 344+CF2​. Hence, CF2≤344+CF2.

 

This inequality simplifies to 344≤CF2, so CF≥18, since CF is positive. Thus, the largest possible value of CF is 18​.

 Dec 10, 2023

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