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In triangle 
, prove that
![\[ a = b\cos C + c\cos B. \]](//latex.artofproblemsolving.com/7/e/d/7ed699f18ede191e5b89bc79d7f0592294347818.png)
, prove that![\[ a = b\cos C + c\cos B. \]](http://latex.artofproblemsolving.com/7/e/d/7ed699f18ede191e5b89bc79d7f0592294347818.png)
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ISI UGB 2025 P7
SomeonecoolLovesMaths 7
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by SomeonecoolLovesMaths
Source: ISI UGB 2025 P7
Consider a ball that moves inside an acute-angled triangle along a straight line, unit it hits the boundary, which is when it changes direction according to the mirror law, just like a ray of light (angle of incidence = angle of reflection). Prove that there exists a triangular periodic path for the ball, as pictured below.
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A Projection Theorem
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Source: VN Math Olympiad For High School Students P1 - 2025
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Y by buratinogigle
too lazy to pull up a diagram
but you consider the cases where the triangle is acute, and where either angle B or angle C is obtuse
if the triangle is acute, dropping an altitude from A basically directly shows it
if the triangle is obtuse (let's say C is, other case basically the same), let the intersection of the altitude from A and the line BC be D, BC=c*cos(B), CD=b*cos(pi-C)=-b*cos(C), a=BC=BD-CD=c*cos(B)-(-b*cos(C))=c*cos(B)+b*cos(C)
but you consider the cases where the triangle is acute, and where either angle B or angle C is obtuse
if the triangle is acute, dropping an altitude from A basically directly shows it
if the triangle is obtuse (let's say C is, other case basically the same), let the intersection of the altitude from A and the line BC be D, BC=c*cos(B), CD=b*cos(pi-C)=-b*cos(C), a=BC=BD-CD=c*cos(B)-(-b*cos(C))=c*cos(B)+b*cos(C)
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