Difference between revisions of "2015 AMC 10A Problems/Problem 19"
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<math> \textbf{(A) }\dfrac{5\sqrt{2}}{3}\qquad\textbf{(B) }\dfrac{50\sqrt{3}-75}{4}\qquad\textbf{(C) }\dfrac{15\sqrt{3}}{8}\qquad\textbf{(D) }\dfrac{50-25\sqrt{3}}{2}\qquad\textbf{(E) }\dfrac{25}{6} </math> | <math> \textbf{(A) }\dfrac{5\sqrt{2}}{3}\qquad\textbf{(B) }\dfrac{50\sqrt{3}-75}{4}\qquad\textbf{(C) }\dfrac{15\sqrt{3}}{8}\qquad\textbf{(D) }\dfrac{50-25\sqrt{3}}{2}\qquad\textbf{(E) }\dfrac{25}{6} </math> | ||
− | ==Solution 1== | + | ==Solution 1 (No Trigonometry) == |
+ | |||
+ | [[File:2015AMC10AProblem19Picture.png]] | ||
<math>\triangle ADC</math> can be split into a <math>45-45-90</math> right triangle and a <math>30-60-90</math> right triangle by dropping a perpendicular from <math>D</math> to side <math>AC</math>. Let <math>F</math> be where that perpendicular intersects <math>AC</math>. | <math>\triangle ADC</math> can be split into a <math>45-45-90</math> right triangle and a <math>30-60-90</math> right triangle by dropping a perpendicular from <math>D</math> to side <math>AC</math>. Let <math>F</math> be where that perpendicular intersects <math>AC</math>. | ||
Line 11: | Line 13: | ||
Because the side lengths of a <math>45-45-90</math> right triangle are in ratio <math>a:a:a\sqrt{2}</math>, <math>DF = AF</math>. | Because the side lengths of a <math>45-45-90</math> right triangle are in ratio <math>a:a:a\sqrt{2}</math>, <math>DF = AF</math>. | ||
− | Because the side lengths of a <math>30-60-90</math> right triangle are in ratio <math>a:a\sqrt{3}:2a</math> and <math>AF | + | Because the side lengths of a <math>30-60-90</math> right triangle are in ratio <math>a:a\sqrt{3}:2a</math> and <math>AF + FC = 5</math>, <math>DF = \frac{5 - AF}{\sqrt{3}}</math>. |
Setting the two equations for <math>DF</math> equal to each other, <math>AF = \frac{5 - AF}{\sqrt{3}}</math>. | Setting the two equations for <math>DF</math> equal to each other, <math>AF = \frac{5 - AF}{\sqrt{3}}</math>. | ||
Line 17: | Line 19: | ||
Solving gives <math>AF = DF = \frac{5\sqrt{3} - 5}{2}</math>. | Solving gives <math>AF = DF = \frac{5\sqrt{3} - 5}{2}</math>. | ||
− | The area of <math>\triangle ADC = \frac12 \cdot DF \cdot AC = \frac{25\sqrt{3} - 25}{4}</math>. | + | The area of <math>\triangle ADC =\frac12 \cdot DF \cdot AC = \frac{25\sqrt{3} - 25}{4}</math>. |
<math>\triangle ADC</math> is congruent to <math>\triangle BEC</math>, so their areas are equal. | <math>\triangle ADC</math> is congruent to <math>\triangle BEC</math>, so their areas are equal. | ||
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Solving gives <math>[CDE] = \frac{50 - 25\sqrt{3}}{2}</math>, so the answer is <math>\boxed{\textbf{(D) }\frac{50 - 25\sqrt{3}}{2}}</math> | Solving gives <math>[CDE] = \frac{50 - 25\sqrt{3}}{2}</math>, so the answer is <math>\boxed{\textbf{(D) }\frac{50 - 25\sqrt{3}}{2}}</math> | ||
− | ==Solution 2== | + | ===Note=== |
+ | Another way to get <math>DF</math> is that you assume <math>AF=DF</math> to be equal to <math>a</math>, as previously mentioned, and <math>CF</math> is equal to <math>a\sqrt{3}</math>. <math>AF+DF=5=a+a\sqrt{3}</math> | ||
+ | |||
+ | ==Solution 2 (Trigonometry)== | ||
The area of <math>\triangle ABC</math> is <math>12.5</math>, and so the leg length of <math>45 - 45 - 90</math> <math>\triangle ABC</math> is <math>5.</math> Thus, the altitude to hypotenuse <math>AB</math>, <math>CF</math>, has length <math>\dfrac{5}{\sqrt{2}}</math> by <math>45 - 45 - 90</math> right triangles. Now, it is clear that <math>\angle{ACD} = \angle{BCE} = 30^\circ</math>, and so by the Exterior Angle Theorem, <math>\triangle{CDE}</math> is an isosceles <math>30 - 75 - 75</math> triangle. Thus, <math>DF = CF \tan 15^\circ = \dfrac{5}{\sqrt{2}} (2 - \sqrt{3})</math> by the Half-Angle formula, and so the area of <math>\triangle CDE</math> is <math>DF \cdot CF = \dfrac{25}{2} (2 - \sqrt{3})</math>. The answer is thus <math>\boxed{\textbf{(D) } \frac{50 - 25\sqrt{3}}{2}}</math> | The area of <math>\triangle ABC</math> is <math>12.5</math>, and so the leg length of <math>45 - 45 - 90</math> <math>\triangle ABC</math> is <math>5.</math> Thus, the altitude to hypotenuse <math>AB</math>, <math>CF</math>, has length <math>\dfrac{5}{\sqrt{2}}</math> by <math>45 - 45 - 90</math> right triangles. Now, it is clear that <math>\angle{ACD} = \angle{BCE} = 30^\circ</math>, and so by the Exterior Angle Theorem, <math>\triangle{CDE}</math> is an isosceles <math>30 - 75 - 75</math> triangle. Thus, <math>DF = CF \tan 15^\circ = \dfrac{5}{\sqrt{2}} (2 - \sqrt{3})</math> by the Half-Angle formula, and so the area of <math>\triangle CDE</math> is <math>DF \cdot CF = \dfrac{25}{2} (2 - \sqrt{3})</math>. The answer is thus <math>\boxed{\textbf{(D) } \frac{50 - 25\sqrt{3}}{2}}</math> | ||
− | ==Solution 3== | + | ==Solution 3 (Analytical Geometry)== |
− | Because the area of triangle <math>ABC</math> is <math>12.5</math>, and the triangle is right and isosceles, we can quickly see that the leg length of the triangle <math>ABC</math> is 5. If we put the triangle on the coordinate plane, with vertex <math>C</math> 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, <math>D</math>, and <math>E</math>. Then, you can use the distance formula to get the length of <math>DE</math>. The height is just <math>\frac{5}{\sqrt{2}}</math>, so the area is just <math>DE | + | Because the area of triangle <math>ABC</math> is <math>12.5</math>, and the triangle is right and isosceles, we can quickly see that the leg length of the triangle <math>ABC</math> is 5. If we put the triangle on the coordinate plane, with vertex <math>C</math> 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, <math>D</math>, and <math>E</math>. Then, you can use the distance formula to get the length of <math>DE</math>. The height is just <math>\frac{5}{\sqrt{2}}</math>, so the area is just <math>DE \cdot \frac{5}{\sqrt{2}} \cdot \frac{1}{2}=\boxed{\textbf{(D) } \frac{50 - 25\sqrt{3}}{2}}</math> |
+ | |||
+ | |||
+ | ==Solution 4 (Trigonometry)== | ||
+ | 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</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> | ||
+ | |||
+ | ==Solution 5 (Basic Trigonometry)== | ||
+ | |||
+ | Prerequisite knowledge for this solution: the side ratios of a 30-60-90, and 45-45-90 right triangle. | ||
+ | |||
+ | |||
+ | We let point C be the origin. Since <math>\overline{CD}</math> and <math>\overline{CE}</math> trisect <math>\angle ACB = 90^{\circ}</math>, this means <math>m\angle CEB = 30^{\circ}</math> and the equation of <math>\overline{CE}</math> is <math>y=\frac{\sqrt{3}}{3}</math> (you can figure out that the tangent of 30 degrees gives <math>\frac{\sqrt{3}}{3}</math>). Next, we can find A to be at <math>(0, 5)</math> and B at <math>(5, 0)</math>, so the equation of <math>\overline{AB}</math> is <math>y=-x+5</math>. So we have the system: | ||
+ | |||
+ | <cmath>\begin{cases}y=\frac{\sqrt{3}}{3}x\\y=-x+5\end{cases}</cmath> | ||
+ | |||
+ | By substituting values, we can arrive at <math>\frac{3+\sqrt{3}}{3}x=5</math>, or <math>x=5\cdot\frac{3}{3+\sqrt{3}}=\frac{15}{3+\sqrt{3}}</math>. We multiply <math>x=\frac{15}{3+\sqrt{3}}\cdot\frac{3-\sqrt{3}}{3-\sqrt{3}}=\frac{45-15\sqrt{3}}{6}=\frac{15-5\sqrt{3}}{2}</math>. | ||
+ | |||
+ | Dropping an altitude from E onto <math>\overline{CB}</math>, and calling the intersection point G, we find that <math>\triangle EGB</math> is a 45-45-90 triangle with a leg of <math>\frac{15-5\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{3}=\frac{15\sqrt{3}-15}{6}=\frac{5\sqrt{3}-5}{2}</math>. Thus, <math>EB=\frac{5\sqrt{3}-5}{2}\sqrt{2}=\frac{5\sqrt{6}-5\sqrt{2}}{2}</math>. | ||
+ | |||
+ | Dropping an altitude from C onto <math>\overline{AB}</math>, and calling the intersection point H, we find that <math>CH=\frac{5\sqrt{2}}{2}=BH</math>, and by the theorem of betweenness applied to H, E, and B, we get <math>HE=HB-EB=\frac{5\sqrt{2}}{2}-\frac{5\sqrt{6}-5\sqrt{2}}{2}=\frac{10\sqrt{2}-5\sqrt{6}}{2}</math>. | ||
+ | |||
+ | We are almost done. By symmetry, <math>HD=HE</math>, so to find the area of the triangle CED, we only need to multiply HE by CH, <math>\frac{10\sqrt{2}-5\sqrt{6}}{2}\cdot\frac{5\sqrt{2}}{2}=\frac{100-50\sqrt{3}}{4}=\frac{50-25\sqrt{3}}{2}</math>. This is answer choice <math>\boxed{\textbf{(D) } \frac{50-25\sqrt{3}}{2}}</math> | ||
+ | ~JH. L | ||
+ | ==Solution 6 (Law of Sines)== | ||
+ | We know that the area of the right triangle is <math>\frac{25}{2}</math> and that the two legs are equal, so we can easily tell that the length of the two legs is <math>5</math>. Thus, the hypotenuse <math>AB = 5\sqrt{2}</math> and <math>\angle{CAB} = \angle{ABC} = 45^{\circ}.</math> | ||
+ | Let's quickly define <math>F</math> as the point that bisects <math>\angle{ACB}</math> and <math>\overline{AB}</math>. Then, we can say that the area of the desired triangle is <math>DF \cdot AF</math>. | ||
+ | Let <math>\overline{AD} = \overline{BE} = n.</math> Since <math>D</math> is one of the trisecting points of <math>\angle{ACB}, \angle{ACD} = 30^{\circ}.</math> Because | ||
+ | <cmath>\angle{ADC} = 180^{\circ}-\angle{ACD}-\angle{CAB},</cmath> | ||
+ | <cmath>\angle{ADC} = 180^{\circ} - 45^{\circ} - 30^{\circ} = 105^{\circ}.</cmath> | ||
+ | Now, we can employ the Law of Sines. It tells us that <math>\frac{\sin(\angle{ADC})}{5} = \frac{\sin(\angle{ACD})}{n}</math>. Plugging in our angle values, we get that <cmath>\frac{\sin(105^{\circ})}{5} = \frac{\sin(30^{\circ})}{n}.</cmath> It's easy to find that <math>\sin(105^{\circ}) = \sin(75^{\circ}) = \frac{\sqrt{2}+\sqrt{6}}{4},</math> and that <math>\sin(30^{\circ}) = \frac{1}{2}</math>. Plugging in these values into our previous equation, we get <cmath>\frac{\sqrt{2}+\sqrt{6}}{20} = \frac{1}{2n}.</cmath> Cross multiplying gets us <cmath>2n\sqrt{2}+2n\sqrt{6} = 20,</cmath> and then we simplify like so: <cmath>2n(\sqrt{2}+\sqrt{6}) = 20\rightarrow</cmath> | ||
+ | <cmath>n(\sqrt{2}+\sqrt{6}) = 10\rightarrow</cmath> | ||
+ | <cmath>n = \frac{10}{\sqrt{2}+\sqrt{6}}.</cmath> | ||
+ | Now, using our definition of <math>n</math>, we know that <math>DF</math> = <math>\frac{AB}{2} - n = \frac{5\sqrt{2}}{2} - \frac{10}{\sqrt{2}+\sqrt{6}}</math>. We want to put this under one common denominator, which is pretty simple to execute. That leaves us with <cmath>\frac{5\sqrt{2} \cdot \sqrt{2}+5\sqrt{2} \cdot \sqrt{6} - 20}{2\sqrt{2} + 2\sqrt{6}}=</cmath> | ||
+ | <cmath>\frac{10+10\sqrt{3} - 20}{2\sqrt{2} + 2\sqrt{6}}=</cmath> | ||
+ | <cmath>\frac{10\sqrt{3} - 10}{2(\sqrt{2} + \sqrt{6})}=</cmath> | ||
+ | <cmath>\frac{2(5\sqrt{3} - 5)}{2(\sqrt{2} + \sqrt{6})}=</cmath> | ||
+ | <cmath>\frac{5\sqrt{3} - 5}{\sqrt{2} + \sqrt{6}}.</cmath> | ||
+ | Whew. That was longer than expected. Anyways, quick inspection tells us that <math>AF = \frac{5\sqrt{2}}{2},</math> so now we just have to do some simplifying to find the desired, <math>[CDE]</math>. Let's do that now. | ||
+ | <cmath>[CDE] = \frac{5\sqrt{3} - 5}{\sqrt{2} + \sqrt{6}} \cdot \frac{5\sqrt{2}}{2}=</cmath> | ||
+ | <cmath>[CDE] = \frac{25\sqrt{6} - 25\sqrt{2}}{2(\sqrt{2} + \sqrt{6})}=</cmath> | ||
+ | <cmath>[CDE] = \frac{25(\sqrt{2} - \sqrt{6})}{2(\sqrt{2} + \sqrt{6})}=</cmath> | ||
+ | (We need to take a quick conjugation break. Note that <math>(\sqrt{2} - \sqrt{6})^2 = 8 - 4\sqrt{3}.</math>) | ||
+ | <cmath>[CDE] = \frac{25(\sqrt{2} - \sqrt{6})}{2(\sqrt{2} + \sqrt{6})} \cdot \frac{(\sqrt{2} - \sqrt{6})}{(\sqrt{2} - \sqrt{6})}=</cmath> | ||
+ | <cmath>[CDE] = \frac{200 - 100\sqrt{3}}{2(4)}=</cmath> | ||
+ | <cmath>[CDE] =\boxed{\frac{50 - 25\sqrt{3}}{2} \text{(D)}.}</cmath> | ||
+ | ~Nickelslordm | ||
+ | |||
+ | |||
+ | ==Video Solution:== | ||
+ | |||
+ | https://www.youtube.com/watch?v=JWMIsCS0Ksk | ||
==See Also== | ==See Also== | ||
+ | |||
{{AMC10 box|year=2015|ab=A|num-b=18|num-a=20}} | {{AMC10 box|year=2015|ab=A|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
[[Category: Introductory Geometry Problems]] | [[Category: Introductory Geometry Problems]] |
Latest revision as of 11:31, 2 June 2024
Contents
Problem
The isosceles right triangle has right angle at
and area
. The rays trisecting
intersect
at
and
. What is the area of
?
Solution 1 (No Trigonometry)
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
Note
Another way to get is that you assume
to be equal to
, as previously mentioned, and
is equal to
.
Solution 2 (Trigonometry)
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 (Analytical Geometry)
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 (Trigonometry)
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,
Solution 5 (Basic Trigonometry)
Prerequisite knowledge for this solution: the side ratios of a 30-60-90, and 45-45-90 right triangle.
We let point C be the origin. Since and
trisect
, this means
and the equation of
is
(you can figure out that the tangent of 30 degrees gives
). Next, we can find A to be at
and B at
, so the equation of
is
. So we have the system:
By substituting values, we can arrive at , or
. We multiply
.
Dropping an altitude from E onto , and calling the intersection point G, we find that
is a 45-45-90 triangle with a leg of
. Thus,
.
Dropping an altitude from C onto , and calling the intersection point H, we find that
, and by the theorem of betweenness applied to H, E, and B, we get
.
We are almost done. By symmetry, , so to find the area of the triangle CED, we only need to multiply HE by CH,
. This is answer choice
~JH. L
Solution 6 (Law of Sines)
We know that the area of the right triangle is and that the two legs are equal, so we can easily tell that the length of the two legs is
. Thus, the hypotenuse
and
Let's quickly define
as the point that bisects
and
. Then, we can say that the area of the desired triangle is
.
Let
Since
is one of the trisecting points of
Because
Now, we can employ the Law of Sines. It tells us that
. Plugging in our angle values, we get that
It's easy to find that
and that
. Plugging in these values into our previous equation, we get
Cross multiplying gets us
and then we simplify like so:
Now, using our definition of
, we know that
=
. We want to put this under one common denominator, which is pretty simple to execute. That leaves us with
Whew. That was longer than expected. Anyways, quick inspection tells us that
so now we just have to do some simplifying to find the desired,
. Let's do that now.
(We need to take a quick conjugation break. Note that
)
~Nickelslordm
Video Solution:
https://www.youtube.com/watch?v=JWMIsCS0Ksk
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.