Difference between revisions of "1998 IMO Problems/Problem 5"
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==Problem== | ==Problem== | ||
| − | Let I be the incenter of triangle ABC. Let the incircle of ABC touch the sides | + | Let <math>I</math> be the incenter of triangle <math>ABC</math>. Let the incircle of <math>ABC</math> touch the sides <math>BC</math>, <math>CA</math>, and <math>AB</math> at <math>K</math>, <math>L</math>, and <math>M</math>, respectively. The line through <math>B</math> parallel to <math>MK</math> meets the lines <math>LM</math> and <math>LK</math> at <math>R</math> and <math>S</math>, respectively. Prove that angle <math>RIS</math> is acute. |
| − | BC, CA, and AB at K, L, and M , respectively. The line through B parallel | ||
| − | to | ||
| − | RIS is acute. | ||
==Solution== | ==Solution== | ||
{{solution}} | {{solution}} | ||
| + | |||
| + | Denote the length of the side MB with x. Denote the angle BKM by a and angle BAC by 2b. Using angle chasing and trigonometry, we can derive: RB = x*cos(b)/cos(a-b); BS = x*cos(a-b)/cos(b); RM = x*sin(a)/cos(a-b); SK = x*sin(a)/cos(b). Let's prove that RI^2 + IS^2 > RS^2 because this will imply that angle RIS is acute. Let's calculate RS^2. It is equal to (RB+BS)^2 = RB^2+BS^2 + 2(x^2). RI^2 - RB^2 = IB^2 (since the angle IBK is equal to 90-a, and angle KBS is equal to a, thus angle IBS is equal to 90 degrees). IS^2-BS^2 = IB^2. So we have to prove that 2*(IB^2)>2*(x^2), and this is true because triangle IMB is a right triangle with IB hypotenuse, so IB > x, as desired. | ||
==See Also== | ==See Also== | ||
{{IMO box|year=1998|num-b=4|num-a=6}} | {{IMO box|year=1998|num-b=4|num-a=6}} | ||
Latest revision as of 09:33, 27 July 2024
Problem
Let 
be the incenter of triangle 
. Let the incircle of touch the sides 
, 
, and 
at 
, 
, and 
, respectively. The line through 
parallel to 
meets the lines 
and 
at 
and 
, respectively. Prove that angle 
is acute.
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
Denote the length of the side MB with x. Denote the angle BKM by a and angle BAC by 2b. Using angle chasing and trigonometry, we can derive: RB = x*cos(b)/cos(a-b); BS = x*cos(a-b)/cos(b); RM = x*sin(a)/cos(a-b); SK = x*sin(a)/cos(b). Let's prove that RI^2 + IS^2 > RS^2 because this will imply that angle RIS is acute. Let's calculate RS^2. It is equal to (RB+BS)^2 = RB^2+BS^2 + 2(x^2). RI^2 - RB^2 = IB^2 (since the angle IBK is equal to 90-a, and angle KBS is equal to a, thus angle IBS is equal to 90 degrees). IS^2-BS^2 = IB^2. So we have to prove that 2*(IB^2)>2*(x^2), and this is true because triangle IMB is a right triangle with IB hypotenuse, so IB > x, as desired.
See Also
| 1998 IMO (Problems) • Resources | ||
| Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
| All IMO Problems and Solutions | ||
