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In triangle $STU$, let $M$ be the midpoint of $\overline{ST},$ and let $N$ be on $\overline{TU}$ such that $\overline{SN}$ is an altitude of triangle $STU$.  If $ST = 8$, $SU = 8$, $TU = 8$, and $\overline{SN}$ and $\overline{UM}$ intersect at $X$, then what is $SX$?

 Feb 9, 2024
 #1
avatar+130466 
+1

 

This one is pretty easy

The triangle is equilateral

The altitude of an equilateral triangle  is a median

And since M is the midpoint of ST, then UM is also a median

Their intersection, X, is the centroid of STU

SN = sqrt [ST^2 - TN^2]  = sqrt [ 8^2 - 4^2]  = sqrt [ 48]  = 4sqrt (3)

 

By a property of centroids ....SX = (2/3)(SN)

 

So

 

SX = (2/3) * 4 (sqrt3)  = 8sqrt (3)  / 3  =  8 / sqrt 3   ≈  4.618

 

cool cool cool

 Feb 10, 2024
 #2
avatar+307 
+1

Just as a side note, I'm not criticizing, but most of the time Vxrtate, you answer the question by rationalizing the denominator. In other words, you should put 833unless it specifically states otherwise.

 Feb 10, 2024

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