Difference between revisions of "2021 April MIMC 10 Problems/Problem 22"
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We can set <math>FG=x</math>, so <math>IF=3-\sqrt{2}-x</math>. Using the similar triangle, <math>\frac{GE}{AI}=\frac{FG}{IF}</math>. Plugging the numbers into the ratio, we can get <math>\frac{3-\sqrt{2}}{2\sqrt{2}-1}=\frac{x}{3-\sqrt{2}-x}</math>. | We can set <math>FG=x</math>, so <math>IF=3-\sqrt{2}-x</math>. Using the similar triangle, <math>\frac{GE}{AI}=\frac{FG}{IF}</math>. Plugging the numbers into the ratio, we can get <math>\frac{3-\sqrt{2}}{2\sqrt{2}-1}=\frac{x}{3-\sqrt{2}-x}</math>. | ||
− | <math> | + | <math>~~~~~~~~~~~~~~~~~~~~\frac{3-\sqrt{2}}{2\sqrt{2}-1}=\frac{x}{3-\sqrt{2}-x}</math> |
<math>~(3-\sqrt{2})(3-\sqrt{2}-x)=x(2\sqrt{2}-1)</math> | <math>~(3-\sqrt{2})(3-\sqrt{2}-x)=x(2\sqrt{2}-1)</math> | ||
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Solve for <math>HG</math>: | Solve for <math>HG</math>: | ||
− | Since angle <math>HEC</math> is <math>90\ | + | Since angle <math>HEC</math> is <math>90^\circ</math> and angle <math>HCE</math> is <math>45^\circ</math>, <math>EC=HE=2\sqrt{2}-1</math>. Since <math>GE=3-\sqrt{2}</math>, <math>HG=2\sqrt{2}-1-(3-\sqrt{2})=3\sqrt{2}-4</math>. |
Finally, we can solve for <math>FH^2</math>, that is, <math>FG^2+HG^2=(\frac{34-23\sqrt{2}}{2})^2+(3\sqrt{2}-4)^2=\frac{1175-830\sqrt{2}}{2}</math>. Therefore, our answer would be <math>1175-830+2+2=\fbox{\textbf{(D)} 349}</math>. | Finally, we can solve for <math>FH^2</math>, that is, <math>FG^2+HG^2=(\frac{34-23\sqrt{2}}{2})^2+(3\sqrt{2}-4)^2=\frac{1175-830\sqrt{2}}{2}</math>. Therefore, our answer would be <math>1175-830+2+2=\fbox{\textbf{(D)} 349}</math>. |
Latest revision as of 14:03, 26 April 2021
In the diagram, is a square with area
.
is a diagonal of square
. Square
has area
. Given that point
bisects line segment
, and
is a line segment. Extend
to meet diagonal
and mark the intersection point
. In addition,
is drawn so that
.
can be represented as
where
are not necessarily distinct integers. Given that
, and
does not have a perfect square factor. Find
.
Solution
To start this problem, we can first observe. Notice that is a right triangle because angle
is supplementary to angle
which is a right angle. Therefore, we just have to solve for the length of side
and
.
Solve for :
Triangles
and
are similar triangles, therefore, we can solve for length
.
. Use the technique of sum of squares and square root disintegration,
. Using the same technique,
.
. Now, we can set up a ratio.
We can set , so
. Using the similar triangle,
. Plugging the numbers into the ratio, we can get
.
Solve for :
Since angle
is
and angle
is
,
. Since
,
.
Finally, we can solve for
, that is,
. Therefore, our answer would be
.