Difference between revisions of "1998 JBMO Problems/Problem 2"
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Thus <math>area</math> of pentagon <math>ABCD</math> = <math>area</math> of <math>ABC</math> + <math>area</math> of <math>AED</math> + <math>area</math> of <math>DAC</math> = <math>1/2 + 1/2 = 1</math> | Thus <math>area</math> of pentagon <math>ABCD</math> = <math>area</math> of <math>ABC</math> + <math>area</math> of <math>AED</math> + <math>area</math> of <math>DAC</math> = <math>1/2 + 1/2 = 1</math> | ||
− | By Kris17 | + | |
+ | |||
+ | By <math>Kris17</math> |
Revision as of 01:20, 29 November 2018
Let
Let angle =
Applying cosine rule to triangle we get:
Substituting we get:
From above,
Thus,
So, of triangle
=
Let be the altitude of triangle DAC from A.
So
This implies .
Since is a cyclic quadrilateral with
, traingle
is congruent to
.
Similarly
is a cyclic quadrilateral and traingle
is congruent to
.
So of triangle
+
of triangle
=
of Triangle
.
Thus
of pentagon
=
of
+
of
+
of
=
By