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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 $ST = 13$, $SU = 14$, $TU = 4$, and $\overline{SN}$ and $\overline{UM}$ intersect at $X$, then what is $UX$?

 Jan 28, 2025
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What is UX?

 

\(14^2-x^2=4^2-(x-13)^2\\ 196-x^2=16-x^2+26x-169\\ 26x=349\\ x_U=13.423\\ y_U=\sqrt{196-13.423^2}\\ y_U=3,978\\ x_M=6.5\\ y_M=0\)

 

\(f_{UT}(x)=\frac{y_U-y_T}{x_U-x_T}\cdot (x -13)=\frac{3.978-0}{13.432-13}\cdot (x-13)\\ f_{UT}(x)=9.2083x-119.7079\\ f_{UM}(x)=\frac{y_U-y_M}{x_U-x_M}\cdot (x-6.5)=\frac{3.978-0}{13.432-6.5}\cdot (x-6.5)\\ f_{UM}(x)=0.5739x-3.73035\\ m_{UT}=9.2083\\ m_{SN}=-\frac{1}{m_{UT}}=-\frac{1}{9,2083}\\ m_{SN}=-0.1086\\ f_{SN}(x)=-0.1086x\)

check!

\(\color{black}The\ point\ N\ is\ not\ on\ the\ section\ \overline{UT}.\ Still,\ carry\ on!\)

\(f_{UX}(x)=f_{UM}(x)=0.5739x-3.73035\\ 0.5739x-3.73035=-0.1086x\\ 0.6825x=3.73035\\ x_X=5.4657\\ y_X=0.5739x-3.73035=0.5739\cdot 5.4657-3.73035\\ y_X=-0.59358 \)

 

\(\overline{UX}=\sqrt{(y_U-y_X)^2+(x_U-x_X)^2}\\ \overline{UX}=\sqrt{(3.978+0.594)^2+(13.423-5.466)^2}\\ \color{blue}\overline{UX}=9.177\)

 

smiley !

 Jan 29, 2025
edited by asinus  Jan 29, 2025
edited by asinus  Jan 29, 2025
edited by asinus  Jan 29, 2025
edited by asinus  Jan 30, 2025

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