Difference between revisions of "2023 SSMO Team Round Problems/Problem 3"
(→Solution) |
(→Solution) |
||
Line 4: | Line 4: | ||
==Solution== | ==Solution== | ||
− | Note that <math>\Delta{ABC}</math> is right with the right angle at <math>B</math>. This means that <math>AC</math> is the diameter of the circle. We can divide quadrilateral <math>ADEF</math> into <math>\Delta{DEF}</math> and <math>\Delta{FAD}</math>, both of which are right triangles. Mark the intersection point between BD and AC as G. We can use the fact that <math>\Delta{ABC}</math> and <math>\Delta{AGB}</math> are similar to find that <math>BG=\frac{20\sqrt{2}}{\sqrt{41}}</math> | + | Note that <math>\Delta{ABC}</math> is right with the right angle at <math>B</math>. This means that <math>AC</math> is the diameter of the circle. We can divide quadrilateral <math>ADEF</math> into <math>\Delta{DEF}</math> and <math>\Delta{FAD}</math>, both of which are right triangles. Mark the intersection point between BD and AC as G. We can use the fact that <math>\Delta{ABC}</math> and <math>\Delta{AGB}</math> are similar to find that <math>BG=\frac{20\sqrt{2}}{\sqrt{41}}</math> |
− | |||
− |
Revision as of 02:10, 3 January 2024
Problem
Let be a triangle such that
and
Let
be the circumcircle of
. Let
be on the circle such that
Let
be the point diametrically opposite of
. Let
be the point diametrically opposite
. Find the area of the quadrilateral
in terms of a mixed number
. Find
.
Solution
Note that is right with the right angle at
. This means that
is the diameter of the circle. We can divide quadrilateral
into
and
, both of which are right triangles. Mark the intersection point between BD and AC as G. We can use the fact that
and
are similar to find that