Difference between revisions of "2015 AMC 10A Problems/Problem 19"
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Just like with Solution 1, we drop a perpendicular from <math>D</math> onto <math>AC</math>, splitting it into a <math>30</math>-<math>60</math>-<math>90</math> triangle and a <math>45</math>-<math>45</math>-<math>90</math> triangle. We find that <math>AF=\frac{5\sqrt{3}-5}{2}</math>. | Just like with Solution 1, we drop a perpendicular from <math>D</math> onto <math>AC</math>, splitting it into a <math>30</math>-<math>60</math>-<math>90</math> triangle and a <math>45</math>-<math>45</math>-<math>90</math> triangle. We find that <math>AF=\frac{5\sqrt{3}-5}{2}</math>. | ||
− | Now, since <math>\triangle AEC\cong \triangle BDC</math> by ASA, <math>CE=CD</math>. Since, <math>DF=\frac{5\sqrt{3}-5}{2}</math>, <math>DC=2\cdot \frac{5\sqrt{3}-5}{2}=5\sqrt{3}-5. By the sine area formula, < | + | Now, since <math>\triangle AEC\cong \triangle BDC</math> by ASA, <math>CE=CD</math>. Since, <math>DF=\frac{5\sqrt{3}-5}{2}</math>, <math>DC=2\cdot \frac{5\sqrt{3}-5}{2}=5\sqrt{3}-5</math>. By the sine area formula, <math>[CDE]=\frac{1}{2}\cdot \sin 30\cdot CD^2=\frac{1}{4}\cdot (100-50\sqrt{3})=\frac{50-25\sqrt{3}}{2}\implies \boxed{\textbf{(D)}</math> |
==See Also== | ==See Also== |
Revision as of 03:37, 25 April 2018
Problem
The isosceles right triangle has right angle at
and area
. The rays trisecting
intersect
at
and
. What is the area of
?
Solution 1
can be split into a
right triangle and a
right triangle by dropping a perpendicular from
to side
. Let
be where that perpendicular intersects
.
Because the side lengths of a right triangle are in ratio
,
.
Because the side lengths of a right triangle are in ratio
and
+
,
.
Setting the two equations for equal to each other,
.
Solving gives .
The area of .
is congruent to
, so their areas are equal.
A triangle's area can be written as the sum of the figures that make it up, so .
.
Solving gives , so the answer is
Solution 2
The area of is
, and so the leg length of
is
Thus, the altitude to hypotenuse
,
, has length
by
right triangles. Now, it is clear that
, and so by the Exterior Angle Theorem,
is an isosceles
triangle. Thus,
by the Half-Angle formula, and so the area of
is
. The answer is thus
Solution 3
Because the area of triangle is
, and the triangle is right and isosceles, we can quickly see that the leg length of the triangle
is 5. If we put the triangle on the coordinate plane, with vertex
at the origin, and the hypotenuse in the first quadrant, we can use slope-intercept form and tangents to get three lines that intersect at the origin,
, and
. Then, you can use the distance formula to get the length of
. The height is just
, so the area is just
Solution 4
Just like with Solution 1, we drop a perpendicular from onto
, splitting it into a
-
-
triangle and a
-
-
triangle. We find that
.
Now, since by ASA,
. Since,
,
. By the sine area formula, $[CDE]=\frac{1}{2}\cdot \sin 30\cdot CD^2=\frac{1}{4}\cdot (100-50\sqrt{3})=\frac{50-25\sqrt{3}}{2}\implies \boxed{\textbf{(D)}$ (Error compiling LaTeX. Unknown error_msg)
See Also
2015 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.