In triangle $ABC$, given lines $l_{b}$ and $l_{c}$ containing the bisectors of angles $B$ and $C$, and the foot $L_{1}$ of the bisector of angle $A$. Restore triangle $ABC$.
Problem
Source: Sharygin Geometry Olympiad 2012 - Problem 9
Tags: geometry, geometric transformation, reflection, geometry unsolved
stats11
30.04.2012 08:09
[asy][asy]
import graph; size(10cm); real labelscalefactor = 0.5; pen dps =
linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; real xmin
= -5.04, xmax = 16.78, ymin = -3.47, ymax = 9.01; draw((xmin, -0.34*xmin +
1.75)--(xmax, -0.34*xmax + 1.75)); draw((xmin, 0.78*xmin + 0.07)--(xmax,
0.78*xmax + 0.07)); draw((xmin, -2.85*xmin + 5.51)--(xmax, -2.85*xmax + 5.51));
draw((xmin, 4.06*xmin + 8.56)--(xmax, 4.06*xmax + 8.56)); draw((xmin,
0*xmin-1.97)--(xmax, 0*xmax-1.97)); draw((xmin, -0.77*xmin + 6.43)--(xmax,
-0.77*xmax + 6.43)); label("$l_{b}$",(5.53,5.26+0.2),SE*labelscalefactor);
label("$l_{c}$",(-3.43,3.68),SE*labelscalefactor);
label("$l_{a}$",(0.06+0.2,5.4),SE*labelscalefactor); dot((2.63,-1.98),dotstyle);
label("$L_1$", (2.75,-1.8), NE * labelscalefactor); dot((1.5,1.24),dotstyle);
label("$I$", (2.08,1.22), NE * labelscalefactor); dot((4.36,3.1),dotstyle);
label("$L_2$", (4.48,3.27-0.3), NE * labelscalefactor); dot((-1.89,0.9),dotstyle);
label("$L_3$", (-2.49-0.5,0.75), NE * labelscalefactor); dot((0.35,-1.97),dotstyle);
label("$L_4$", (-0.03,-1.59), NE * labelscalefactor); dot((2.55,4.48),dotstyle);
label("$L_5$", (2.66,4.64), NE * labelscalefactor); dot((-1.34,3.11),dotstyle);
label("$L_6$", (-1.99-0.3,3.27), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy][/asy]
1. Let $I$ be the intersection of $\ell_{b}$ and $\ell_{c}$.
2. Let $\ell_{a}$ be the line passing through $L_{1}$ and $I$.
3. Let $L_{2}$ be reflection of $L_{1}$ over $\ell_{c}$.
4. Let $L_{3}$ be reflection of $L_{2}$ over $\ell_{a}$.
5. Let $L_{4}$ be reflection of $L_{3}$ over $\ell_{b}$.
6. Let $L_{5}$ be reflection of $L_{4}$ over $\ell_{c}$.
7. The triangle formed by lines $L_{1}L_{4}, L_{2}L_{5}, L_{3}L_{6}$ is what we want.
Vo Duc Dien
09.05.2012 00:52
Another way to solve but you don't want to do is solve $ x $ amount of equations with same amount of unknowns with the length $ L_{1}I $ given by applying the property of the angle bisectors.