Try this beautiful problem from PRMO, 2016 based on Triangle
In a triangle ABC right angled at vertex B, a point O is choosen on the side BC such that the circle Y centerer at O of radius OB touches the side AC.Let AB =63 and BC=16 and the radius of Y be the form mn wherw m,n are relatively prime positive integers,find the value of m+n?
Geometry
Triangle
Circle
But try the problem first...
Answer:\(71\)
PRMO-2016, Problem 10
Pre College Mathematics
First hint
Given that ABC is a triangle where AB=63 and BC=16 and \(\angle ABC=90^{\circ}\). O is any point on BC.Let OB=\(r\)=radius of the circle.we have to find out the value of \(r\) for \(m+n\).
Now \(\triangle ABC\) is aright angle triangle at\(B\) So using pythagoras theorm \(AC=\sqrt {(AB)^2+(BC)^2}=\sqrt{(63)^2+(16)^2}=65\).now \(\triangle ODC\) is also a right angle triangle as OD radius and AC tangent,can you find out the value of \(r\)?
Can you now finish the problem ..........
Second Step
Given bc=16,Let OB=\(r\) then OC=\((16- r)\).we have to find out the value of \(r\).if we can show that \(\triangle ABC \sim \triangle ODC\) then we can find out the value of r.can you proof \(\triangle ABC \sim \triangle ODC\)?
Second Step
In \(\triangle ABC\) & \(\triangle ODC\),
\(\angle ABC=\angle ODC\)
\(\angle C= \angle C\)
Therefore \(\triangle ABC \sim \triangle ODC\)
so we have ....
\(\frac{AC}{OC}=\frac{AB}{OD}=\frac{BC}{AC}\) \(\Rightarrow \frac{65}{16-r}=\frac{63}{r}=\frac{16}{65}\)
Therefore,\(\frac{65}{16-r}=\frac{63}{r}\Rightarrow r=\frac{63}{8}\)=\(\frac{m}{n}\)
Therefore \((m+n)=63+8=71\)
AMC - AIME Boot Camp... for brilliant students.
Use our exclusive one-on-one plus group class system to prepare for Math Olympiad
Try this beautiful problem from PRMO, 2016 based on Triangle
In a triangle ABC right angled at vertex B, a point O is choosen on the side BC such that the circle Y centerer at O of radius OB touches the side AC.Let AB =63 and BC=16 and the radius of Y be the form mn wherw m,n are relatively prime positive integers,find the value of m+n?
Geometry
Triangle
Circle
But try the problem first...
Answer:\(71\)
PRMO-2016, Problem 10
Pre College Mathematics
First hint
Given that ABC is a triangle where AB=63 and BC=16 and \(\angle ABC=90^{\circ}\). O is any point on BC.Let OB=\(r\)=radius of the circle.we have to find out the value of \(r\) for \(m+n\).
Now \(\triangle ABC\) is aright angle triangle at\(B\) So using pythagoras theorm \(AC=\sqrt {(AB)^2+(BC)^2}=\sqrt{(63)^2+(16)^2}=65\).now \(\triangle ODC\) is also a right angle triangle as OD radius and AC tangent,can you find out the value of \(r\)?
Can you now finish the problem ..........
Second Step
Given bc=16,Let OB=\(r\) then OC=\((16- r)\).we have to find out the value of \(r\).if we can show that \(\triangle ABC \sim \triangle ODC\) then we can find out the value of r.can you proof \(\triangle ABC \sim \triangle ODC\)?
Second Step
In \(\triangle ABC\) & \(\triangle ODC\),
\(\angle ABC=\angle ODC\)
\(\angle C= \angle C\)
Therefore \(\triangle ABC \sim \triangle ODC\)
so we have ....
\(\frac{AC}{OC}=\frac{AB}{OD}=\frac{BC}{AC}\) \(\Rightarrow \frac{65}{16-r}=\frac{63}{r}=\frac{16}{65}\)
Therefore,\(\frac{65}{16-r}=\frac{63}{r}\Rightarrow r=\frac{63}{8}\)=\(\frac{m}{n}\)
Therefore \((m+n)=63+8=71\)
AMC - AIME Boot Camp... for brilliant students.
Use our exclusive one-on-one plus group class system to prepare for Math Olympiad
