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 $CF.$

tomtom Dec 10, 2023

#1**0 **

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+21CF2=21AB⋅BC.

This simplifies to 72+200+21CF2=21AB⋅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.

BuiIderBoi Dec 10, 2023