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Let STU be a triangle with side lengths ST=22, TU=8 and SU=22. Let M be the midpoint of ST and let N be on TU such that SN is an altitude of

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Let STU be a triangle with side lengths ST=22, TU=8 and SU=22. Let M be the midpoint of ST and let N be on TU such that SN is an altitude of triangle STU. If SN and UM intersect at X, then what is SX?

Aug 11, 2022

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Note that because $$\triangle STU$$ is isosceles, $$SN$$ is also a median.

Now, draw the median $$TK$$. This intersects at $$X$$, and by the properties of medians, we know that $$SX = {2 \over 3}\times SN$$.

$$\triangle SNU$$ is a right triangle with $$SU = 22$$ and $$NU = 8 \div 2 = 4$$, meaning $$SN = \sqrt{22^2-4^2} = \sqrt{468} = 6 \sqrt{13}$$

So, $$SX = {2 \over 3} \times 6 \sqrt{13} = \color{brown}\boxed{4 \sqrt {13}}$$

Aug 12, 2022