Difference between revisions of "2022 AIME II Problems/Problem 15"
m (→Solution 3) |
(→Problem) |
||
Line 8: | Line 8: | ||
circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B); | circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B); | ||
point A=intersectionpoints(o,w2)[1],D=intersectionpoints(o,w2)[0],C=intersectionpoints(o,w1)[0]; | point A=intersectionpoints(o,w2)[1],D=intersectionpoints(o,w2)[0],C=intersectionpoints(o,w1)[0]; | ||
− | filldraw(A--B--O1--C--D--O2--cycle,0.2* | + | filldraw(A--B--O1--C--D--O2--cycle,0.2*fuchsia+white,black); |
draw(w1); | draw(w1); | ||
draw(w2); | draw(w2); |
Revision as of 16:08, 25 December 2022
Problem
Two externally tangent circles and
have centers
and
, respectively. A third circle
passing through
and
intersects
at
and
and
at
and
, as shown. Suppose that
,
,
, and
is a convex hexagon. Find the area of this hexagon.
Solution 1
First observe that and
. Let points
and
be the reflections of
and
, respectively, about the perpendicular bisector of
. Then quadrilaterals
and
are congruent, so hexagons
and
have the same area. Furthermore, triangles
and
are congruent, so
and quadrilateral
is an isosceles trapezoid.
Next, remark that
, so quadrilateral
is also an isosceles trapezoid; in turn,
, and similarly
. Thus, Ptolmey's theorem on
yields
, whence
. Let
. The Law of Cosines on triangle
yields
and hence
. Thus the distance between bases
and
is
(in fact,
is a
triangle with a
triangle removed), which implies the area of
is
.
Now let and
; the tangency of circles
and
implies
. Furthermore, angles
and
are opposite angles in cyclic quadrilateral
, which implies the measure of angle
is
. Therefore, the Law of Cosines applied to triangle
yields
Thus , and so the area of triangle
is
.
Thus, the area of hexagon is
.
~djmathman
Solution 2
Denote by the center of
.
Denote by
the radius of
.
We have ,
,
,
,
,
are all on circle
.
Denote .
Denote
.
Denote
.
Because and
are on circles
and
,
is a perpendicular bisector of
. Hence,
.
Because and
are on circles
and
,
is a perpendicular bisector of
. Hence,
.
In ,
Hence,
In ,
Hence,
In ,
Hence,
Taking , we get
.
Thus,
.
Taking these into (1), we get .
Hence,
Hence, .
In ,
In , by applying the law of sines, we get
Because circles and
are externally tangent,
is on circle
,
is on circle
,
Thus, .
Now, we compute and
.
Recall and
.
Thus,
.
We also have
Thus,
Therefore,
~Steven Chen (www.professorchenedu.com)
Solution 3
Let points and
be the reflections of
and
respectively, about the perpendicular bisector of
We establish the equality of the arcs and conclude that the corresponding chords are equal
Similarly
Ptolmey's theorem on yields
The area of the trapezoid
is equal to the area of an isosceles triangle with sides
and
The height of this triangle is The area of
is
Denote
hence
Semiperimeter of is
The distance from the vertex to the tangent points of the inscribed circle of the triangle
is equal
The radius of the inscribed circle is
The area of triangle is
The hexagon has the same area as hexagon
The area of hexagon is equal to the sum of the area of the trapezoid
and the areas of two equal triangles
and
so the area of the hexagon
is
vladimir.shelomovskii@gmail.com, vvsss
See Also
2022 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.