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In triangle $PQR,$ let $X$ be the intersection of the angle bisector of $\angle P$ with side $\overline{QR}$, and let $Y$ be the foot of the perpendicular from $X$ to side $\overline{PR}$.  If $PQ = 8,$ $QR = 5,$ and $PR = 1,$ then compute the length of $\overline{XY}$.

 Apr 15, 2024
 #1
avatar+179 
+1

Unfortunately, with the given information (side lengths PQ=8, QR=5, and PR=1), it is impossible to determine the length of XY. Here's why:

Triangle Inequality Violation: The side lengths provided violate the triangle inequality. The triangle inequality states that the sum of any two side lengths in a triangle must be greater than the third side length. In this case, PQ+PR=8+1=9, which is not greater than QR=5. Therefore, a triangle with these side lengths cannot exist.

Ambiguous Angle Bisector: Even if the side lengths were valid, the information provided wouldn't be sufficient to determine the length of XY. The angle bisector of angle P divides angle P into two ratios, which depend on the original measure of angle P. Without knowing the angle measure or additional information about the triangle (like area), it's impossible to determine the specific location of point X on side QR​. Consequently, the distance XY cannot be determined.

 Apr 15, 2024
 #2
avatar+14944 
+1

The question can be solved by specifying PR=10 instead of PR=1.

laugh !

 Apr 16, 2024
edited by asinus  Apr 17, 2024
 #3
avatar+14944 
+1

 PQ = 8, QR = 5, and PR = 10.

 

\(f_{PR}(x)=0\\ 5^2=8^2+10^2-2\cdot 8\cdot 10\cdot cos\ P\\ 160\cdot cos\ \ P=64+100-25\\ cos\ P=\frac{64+100-25}{160}\\ P=29.686° \)

 

\(f_{PX}(x)=tan\ (\frac{29.686}{2})\cdot x\\ f_{PX}(x)=0.265x\\ sin\ R:sin\ P=8:5\\ sin\ \ R= \dfrac{8\cdot sin\ P }{5}=\dfrac{8\cdot sin\ 29.686° }{5}\\ R=52.4105°\\ f_{RQ}(x)=tan\ (-R)\cdot (x-10)\\ f_{RQ}(x)=tan\ (-52,4105°)\cdot (x-10)\\ f_{RQ}(x)=-1,3\cdot (x-10)\\\)

 

\(f_{PX}(x)=f_{PQ}(x)\\ 0.265x=-1.3(x-10)\\ 0.265x=-1.3x+13\\ x=\dfrac{13}{1.565}\\ \color{blue}x_X=8.3067\\ \color{blue}y_X=2.2013\)

 

\(f_{XY}(x)=m(x-x_X)+y_X\\ m=tan\ (90°-R)=tan\ (90°-52.4105°)\\ m=0.7698\\ f_{XY}(x)=0.7698(x-8.3067)+2.2013\\ f_{XY}(x)=0.7698x-6.3946+2.2013=0\\ \color{blue}x_Y=5.4472\\ \color{blue}y_Y=0\)

 

\(L_{XY}=\sqrt{y_X^2+(x_X-x_Y)^2}=\sqrt{2.2013^2+(8.3067-5.4472)^2}\\ \color{blue}L_{XY}=3.6087\)

 

\(\color{blue}The\ length\ of\ \overline{XY}\ is\ 3.6807.\)

 

laugh !

 Apr 16, 2024
edited by asinus  Apr 16, 2024
edited by asinus  Apr 17, 2024
edited by asinus  Apr 17, 2024

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