Difference between revisions of "2021 April MIMC 10 Problems/Problem 23"
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The sum of the area of the regions <math>ABH</math> (formed by side <math>AB</math>, arc <math>BH</math>, and side <math>HA</math>), <math>DFE</math> (formed by side <math>ED</math>, arc <math>DF</math>, and side <math>FE</math>) and sectors <math>FGH</math> and <math>BCD</math> can be calculated by turning <math>AHBO</math> into a square, and subtract the extra areas. Since <math>BO</math> has length <math>10</math>, we know that the height of the two right triangles are <math>2</math> and the based are <math>12</math>. <math>144-24=120</math>. We want to also subtract the shaded quarter circle. The area is <math>120-25\pi</math>. The region enclosed by arc <math>DF</math> and length <math>DE, FE</math> is the reflection of the previous area. The area <math>HGF=120-100+25\pi=20+25\pi</math>. The region <math>BCD</math> is also the reflection. Therefore, the total area is <math>120-25\pi+120-25\pi+20+25\pi+20+25\pi=280</math>. | The sum of the area of the regions <math>ABH</math> (formed by side <math>AB</math>, arc <math>BH</math>, and side <math>HA</math>), <math>DFE</math> (formed by side <math>ED</math>, arc <math>DF</math>, and side <math>FE</math>) and sectors <math>FGH</math> and <math>BCD</math> can be calculated by turning <math>AHBO</math> into a square, and subtract the extra areas. Since <math>BO</math> has length <math>10</math>, we know that the height of the two right triangles are <math>2</math> and the based are <math>12</math>. <math>144-24=120</math>. We want to also subtract the shaded quarter circle. The area is <math>120-25\pi</math>. The region enclosed by arc <math>DF</math> and length <math>DE, FE</math> is the reflection of the previous area. The area <math>HGF=120-100+25\pi=20+25\pi</math>. The region <math>BCD</math> is also the reflection. Therefore, the total area is <math>120-25\pi+120-25\pi+20+25\pi+20+25\pi=280</math>. | ||
− | As a result, the ratio is <math>\ | + | As a result, the ratio is <math>\boxed{\textbf{(A)} \frac{50\pi+1}{280}}</math>. |
Latest revision as of 14:04, 26 April 2021
On a coordinate plane, point denotes the origin which is the center of the diamond shape in the middle of the figure. Point
has coordinate
, and point
,
, and
are formed through
,
, and
rotation about the origin
, respectively. Quarter circle
(formed by the arc
and line segments
and
) has area
. Furthermore, another quarter circle
formed by arc
and line segments
,
is formed through a reflection of sector
across the line
. The small diamond centered at
is a square, and the area of the little square is
. Let
denote the area of the shaded region, and
denote the sum of the area of the regions
(formed by side
, arc
, and side
),
(formed by side
, arc
, and side
) and sectors
and
. Find
in the simplest radical form.
Solution
First of all, we know that . Since the area of the quarter circle is
, we can get that
Then, we can calculate the area of shaded region. It is made of two quarter circles and two right triangles. The total area would be
.
The sum of the area of the regions (formed by side
, arc
, and side
),
(formed by side
, arc
, and side
) and sectors
and
can be calculated by turning
into a square, and subtract the extra areas. Since
has length
, we know that the height of the two right triangles are
and the based are
.
. We want to also subtract the shaded quarter circle. The area is
. The region enclosed by arc
and length
is the reflection of the previous area. The area
. The region
is also the reflection. Therefore, the total area is
.
As a result, the ratio is .