Difference between revisions of "2023 SSMO Team Round Problems/Problem 3"
(→Solution) |
(→Solution) |
||
(6 intermediate revisions by the same user not shown) | |||
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} and \Delta{FAD}</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>, so <math>BD=\frac{40\sqrt{2}}{\sqrt{41}}=FE</math> by symmetry. Then, <math>FD=\frac{9\sqrt{2}}{\sqrt{41}}=FE</math> by the Pythagorean Theorem, so the area of <math>\Delta{DEF}</math> is <math>\frac{360}{41}</math>. Since <math>DA=FA=4\sqrt{2}</math> by symmetry, <math>FA=4\sqrt{2}</math>, so the area of <math>\Delta{FAD}</math> is <math>20</math>. This means that the area of the entire quadrilateral equals <math>\frac{360}{41}+20=28\frac{32}{41}</math>, so the answer is <math>28+32+41=\boxed{101}</math>. |
+ | |||
+ | ~alexanderruan |
Latest 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
, so
by symmetry. Then,
by the Pythagorean Theorem, so the area of
is
. Since
by symmetry,
, so the area of
is
. This means that the area of the entire quadrilateral equals
, so the answer is
.
~alexanderruan