Difference between revisions of "2016 AMC 10A Problems/Problem 24"
m (→Solution 8 (Cheap Solution - it doesn't make any cents!)) |
m |
||
Line 4: | Line 4: | ||
<math>\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500</math> | <math>\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500</math> | ||
− | + | ==Solution 1 (Algebra)== | |
To save us from getting big numbers with lots of zeros behind them, let's divide all side lengths by <math>200</math> for now, then multiply it back at the end of our solution. | To save us from getting big numbers with lots of zeros behind them, let's divide all side lengths by <math>200</math> for now, then multiply it back at the end of our solution. | ||
Line 71: | Line 71: | ||
Finally, we multiply back the <math>200</math> that we divided by at the beginning of the problem to get <math>AD=\boxed{500 (E)}</math>. | Finally, we multiply back the <math>200</math> that we divided by at the beginning of the problem to get <math>AD=\boxed{500 (E)}</math>. | ||
− | + | ==Solution 2 (Trigonometry Bash)== | |
<asy> | <asy> | ||
size(250); | size(250); | ||
Line 113: | Line 113: | ||
Because <math>\angle AOB</math>, <math>\angle BOC</math>, and <math>\angle COD</math> are congruent, <math>\angle AOD = 3\theta</math>. To find the remaining side (<math>AD</math>), we simply have to apply the law of cosines to <math>\Delta AOD</math> . Now, to find <math>\cos 3\theta</math>, we can derive a formula that only uses <math>\cos\theta</math>: <cmath>\cos 3\theta=\cos (2\theta+\theta)= \cos 2\theta \cos\theta- \sin 2\theta \cdot (2\sin\theta \cos\theta)</cmath> <cmath>\cos 3\theta= \cos\theta (\cos 2\theta-2\sin^{2}\theta)=\cos\theta (2\cos^{2}\theta-1+2\cos^{2}\theta)</cmath> <cmath>\Rightarrow \cos 3\theta=4\cos^{3}\theta-3\cos\theta</cmath> Plugging in <math>\cos\theta=\frac{3}{4}</math>, we get <math>\cos 3\theta= -\frac{9}{16}</math>. Now, applying law of cosines on triangle <math>OAD</math>, we get <cmath>(AD)^{2}= 2\cdot (200)^{2}+ 2\cdot (200)^{2}+2\cdot 200\sqrt2 \cdot 200\sqrt2 \cdot \frac{9}{16}</cmath> <cmath>\Rightarrow 2\cdot (200)^{2} \cdot (1+1+ \frac{9}{8})=(200)^{2}\cdot \frac{25}{4}</cmath> <cmath>AD=200 \cdot \frac{5}{2}=\boxed{500}</cmath> | Because <math>\angle AOB</math>, <math>\angle BOC</math>, and <math>\angle COD</math> are congruent, <math>\angle AOD = 3\theta</math>. To find the remaining side (<math>AD</math>), we simply have to apply the law of cosines to <math>\Delta AOD</math> . Now, to find <math>\cos 3\theta</math>, we can derive a formula that only uses <math>\cos\theta</math>: <cmath>\cos 3\theta=\cos (2\theta+\theta)= \cos 2\theta \cos\theta- \sin 2\theta \cdot (2\sin\theta \cos\theta)</cmath> <cmath>\cos 3\theta= \cos\theta (\cos 2\theta-2\sin^{2}\theta)=\cos\theta (2\cos^{2}\theta-1+2\cos^{2}\theta)</cmath> <cmath>\Rightarrow \cos 3\theta=4\cos^{3}\theta-3\cos\theta</cmath> Plugging in <math>\cos\theta=\frac{3}{4}</math>, we get <math>\cos 3\theta= -\frac{9}{16}</math>. Now, applying law of cosines on triangle <math>OAD</math>, we get <cmath>(AD)^{2}= 2\cdot (200)^{2}+ 2\cdot (200)^{2}+2\cdot 200\sqrt2 \cdot 200\sqrt2 \cdot \frac{9}{16}</cmath> <cmath>\Rightarrow 2\cdot (200)^{2} \cdot (1+1+ \frac{9}{8})=(200)^{2}\cdot \frac{25}{4}</cmath> <cmath>AD=200 \cdot \frac{5}{2}=\boxed{500}</cmath> | ||
− | + | ==Solution 3 (Easier trig)== | |
<asy> | <asy> | ||
Line 169: | Line 169: | ||
Thus, we have <math>\frac{FC}{DC} = \cos \theta = \frac{3}{4}</math>, so <math>FC = 150</math>. Similarly, <math>BE = 150</math>, and <math>AD = 150 + 200 + 150 = \boxed{500}</math>. | Thus, we have <math>\frac{FC}{DC} = \cos \theta = \frac{3}{4}</math>, so <math>FC = 150</math>. Similarly, <math>BE = 150</math>, and <math>AD = 150 + 200 + 150 = \boxed{500}</math>. | ||
− | + | ==Solution 4 (Just Geometry)== | |
<asy> | <asy> | ||
Line 236: | Line 236: | ||
<math>OE = 100\sqrt{2} = \frac{OB}{2}</math> and <math>EF=\frac{BC}{2}=100</math> , so <math>AD=AE + EF + FD = 200 + 100 + 200 = \boxed{\textbf{(E) } 500}</math> | <math>OE = 100\sqrt{2} = \frac{OB}{2}</math> and <math>EF=\frac{BC}{2}=100</math> , so <math>AD=AE + EF + FD = 200 + 100 + 200 = \boxed{\textbf{(E) } 500}</math> | ||
− | + | ==Solution 5 (Ptolemy's Theorem)== | |
<asy> | <asy> | ||
Line 258: | Line 258: | ||
Now by Ptolemy's Theorem we have <math>CA^2 = s^2 + AD \cdot s \implies AD = (7/2-1)s.</math> This gives us: <cmath>\boxed{\textbf{(E) } 500.}</cmath> | Now by Ptolemy's Theorem we have <math>CA^2 = s^2 + AD \cdot s \implies AD = (7/2-1)s.</math> This gives us: <cmath>\boxed{\textbf{(E) } 500.}</cmath> | ||
− | + | ==Solution 6 (Trigonometry)== | |
Since all three sides equal <math>200</math>, they subtend three equal angles from the center. The right triangle between the center of the circle, a vertex, and the midpoint between two vertices has side lengths <math>100,100\sqrt{7},200\sqrt{2}</math> by the Pythagorean Theorem. Thus, the sine of half of the subtended angle is <math>\frac{100}{200\sqrt{2}}=\frac{\sqrt{2}}{4}</math>. Similarly, the cosine is <math>\frac{100\sqrt{7}}{200\sqrt{2}}=\frac{\sqrt{14}}{4}</math>. | Since all three sides equal <math>200</math>, they subtend three equal angles from the center. The right triangle between the center of the circle, a vertex, and the midpoint between two vertices has side lengths <math>100,100\sqrt{7},200\sqrt{2}</math> by the Pythagorean Theorem. Thus, the sine of half of the subtended angle is <math>\frac{100}{200\sqrt{2}}=\frac{\sqrt{2}}{4}</math>. Similarly, the cosine is <math>\frac{100\sqrt{7}}{200\sqrt{2}}=\frac{\sqrt{14}}{4}</math>. | ||
Since there are three sides, and since <math>\sin\theta=\sin\left(180-\theta\right)</math>,we seek to find <math>2r\sin 3\theta</math>. | Since there are three sides, and since <math>\sin\theta=\sin\left(180-\theta\right)</math>,we seek to find <math>2r\sin 3\theta</math>. | ||
Line 265: | Line 265: | ||
<cmath>2r\sin 3\theta=2\left(200\sqrt{2}\right)\left(\frac{5\sqrt{2}}{8}\right)=400\sqrt{2}\left(\frac{5\sqrt{2}}{8}\right)=\frac{800\cdot 5}{8}=\boxed{\textbf{(E)}\text{ 500}}</cmath> | <cmath>2r\sin 3\theta=2\left(200\sqrt{2}\right)\left(\frac{5\sqrt{2}}{8}\right)=400\sqrt{2}\left(\frac{5\sqrt{2}}{8}\right)=\frac{800\cdot 5}{8}=\boxed{\textbf{(E)}\text{ 500}}</cmath> | ||
− | + | ==Solution 7 (Area)== | |
For simplicity, scale everything down by a factor of 100. Let the inscribed trapezoid be <math>ABCD</math>, where <math>AB=BC=CD=2</math> and <math>DA</math> is the missing side length. Let <math>DA=2x</math>. If <math>M</math> and <math>N</math> are the midpoints of <math>BC</math> and <math>AD</math>, respectively, the height of the trapezoid is <math>OM-ON</math>. By the pythagorean theorem, <math>OM=\sqrt{OB^2-BM^2}=\sqrt7</math> and <math>ON=\sqrt{OA^2-AN^2}=\sqrt{8-x^2}</math>. Thus the height of the trapezoid is <math>\sqrt7-\sqrt{8-x^2}</math>, so the area is <math>\frac{(2+2x)(\sqrt7-\sqrt{8-x^2})}{2}=(x+1)(\sqrt7-\sqrt{8-x^2})</math>. By Brahmagupta's formula, the area is <math>\sqrt{(x+1)(x+1)(x+1)(3-x)}</math>. Setting these two equal, we get <math>(x+1)(\sqrt7-\sqrt{8-x^2})=\sqrt{(x+1)(x+1)(x+1)(3-x)}</math>. Dividing both sides by <math>x+1</math> and then squaring, we get <math>7-2(\sqrt7)(\sqrt{8-x^2})+8-x^2=(x+1)(3-x)</math>. Expanding the right hand side and canceling the <math>x^2</math> terms gives us <math>15-2(\sqrt7)(\sqrt{8-x^2})=2x+3</math>. Rearranging and dividing by two, we get <math>(\sqrt7)(\sqrt{8-x^2})=6-x</math>. Squaring both sides, we get <math>56-7x^2=x^2-12x+36</math>. Rearranging, we get <math>8x^2-12x-20=0</math>. Dividing by 4 we get <math>2x^2-3x-5=0</math>. Factoring we get, <math>(2x-5)(x+1)=0</math>, and since <math>x</math> cannot be negative, we get <math>x=2.5</math>. Since <math>DA=2x</math>, <math>DA=5</math>. Scaling up by 100, we get <math>\boxed{\textbf{(E)}\text{ 500}}</math>. | For simplicity, scale everything down by a factor of 100. Let the inscribed trapezoid be <math>ABCD</math>, where <math>AB=BC=CD=2</math> and <math>DA</math> is the missing side length. Let <math>DA=2x</math>. If <math>M</math> and <math>N</math> are the midpoints of <math>BC</math> and <math>AD</math>, respectively, the height of the trapezoid is <math>OM-ON</math>. By the pythagorean theorem, <math>OM=\sqrt{OB^2-BM^2}=\sqrt7</math> and <math>ON=\sqrt{OA^2-AN^2}=\sqrt{8-x^2}</math>. Thus the height of the trapezoid is <math>\sqrt7-\sqrt{8-x^2}</math>, so the area is <math>\frac{(2+2x)(\sqrt7-\sqrt{8-x^2})}{2}=(x+1)(\sqrt7-\sqrt{8-x^2})</math>. By Brahmagupta's formula, the area is <math>\sqrt{(x+1)(x+1)(x+1)(3-x)}</math>. Setting these two equal, we get <math>(x+1)(\sqrt7-\sqrt{8-x^2})=\sqrt{(x+1)(x+1)(x+1)(3-x)}</math>. Dividing both sides by <math>x+1</math> and then squaring, we get <math>7-2(\sqrt7)(\sqrt{8-x^2})+8-x^2=(x+1)(3-x)</math>. Expanding the right hand side and canceling the <math>x^2</math> terms gives us <math>15-2(\sqrt7)(\sqrt{8-x^2})=2x+3</math>. Rearranging and dividing by two, we get <math>(\sqrt7)(\sqrt{8-x^2})=6-x</math>. Squaring both sides, we get <math>56-7x^2=x^2-12x+36</math>. Rearranging, we get <math>8x^2-12x-20=0</math>. Dividing by 4 we get <math>2x^2-3x-5=0</math>. Factoring we get, <math>(2x-5)(x+1)=0</math>, and since <math>x</math> cannot be negative, we get <math>x=2.5</math>. Since <math>DA=2x</math>, <math>DA=5</math>. Scaling up by 100, we get <math>\boxed{\textbf{(E)}\text{ 500}}</math>. | ||
− | + | ==Solution 8 (Cheap Solution - it doesn't make any cents!)== | |
WLOG, let <math>AB=BC=CD=200</math>, and let ABCD be inscribed in a clrcle with radius <math>200\sqrt2</math>. We draw perpendiculars from <math>B</math> and <math>C</math> to <math>AD</math>, and label the intersections <math>E</math> and <math>F</math>, respectively. We can see that <math>EF=200</math> (because BCFE is a rectangle), and since <math>AD</math> is clearly greater than 200, and and since <math>EF</math>, which is part of segment <math>AD</math>, is an integer, than we conclude that <math>AD</math> is also an integer or of the form <math>200+2*AE</math>. There is no reason for <math>AE</math> to be of the form <math>a\sqrt{b} - 100</math> because it seems too arbitrary. The only other integer choice is <math>\boxed{\textbf{(E)}\text{ 500}}</math>. | WLOG, let <math>AB=BC=CD=200</math>, and let ABCD be inscribed in a clrcle with radius <math>200\sqrt2</math>. We draw perpendiculars from <math>B</math> and <math>C</math> to <math>AD</math>, and label the intersections <math>E</math> and <math>F</math>, respectively. We can see that <math>EF=200</math> (because BCFE is a rectangle), and since <math>AD</math> is clearly greater than 200, and and since <math>EF</math>, which is part of segment <math>AD</math>, is an integer, than we conclude that <math>AD</math> is also an integer or of the form <math>200+2*AE</math>. There is no reason for <math>AE</math> to be of the form <math>a\sqrt{b} - 100</math> because it seems too arbitrary. The only other integer choice is <math>\boxed{\textbf{(E)}\text{ 500}}</math>. |
Revision as of 14:25, 9 February 2017
Contents
Problem
A quadrilateral is inscribed in a circle of radius . Three of the sides of this quadrilateral have length
. What is the length of the fourth side?
Solution 1 (Algebra)
To save us from getting big numbers with lots of zeros behind them, let's divide all side lengths by for now, then multiply it back at the end of our solution.
Construct quadrilateral on the circle with
being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center
to
and
. Let the intersection of
and
be point
. Notice that
and
are perpendicular because
is a kite.
We set lengths equal to
. By the Pythagorean Theorem,
We solve for :
By Ptolemy's Theorem,
Substituting values,
Finally, we multiply back the that we divided by at the beginning of the problem to get
.
Solution 2 (Trigonometry Bash)
Construct quadrilateral on the circle with
being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center
to
and
. Apply law of cosines on
; let
. We get the following equation:
Substituting the values in, we get
Canceling out, we get
Because
,
, and
are congruent,
. To find the remaining side (
), we simply have to apply the law of cosines to
. Now, to find
, we can derive a formula that only uses
:
Plugging in
, we get
. Now, applying law of cosines on triangle
, we get
Solution 3 (Easier trig)
Construct quadrilateral on the circle
with
being the missing side. Then, drop perpendiculars from
and
to (extended) line
, and let these points be
and
, respectively. Also, let
. From Law of Cosines on
, we have
.
Now, since
is isosceles with
, we have that
. By SSS congruence, we have that
, so we have that
, so
.
Thus, we have
, so
. Similarly,
, and
.
Solution 4 (Just Geometry)
Label AD intercept OB at E and OC at F.
From there, , thus:
because they are both radii of
. Since
, we have that
. Similarly,
.
and
, so
Solution 5 (Ptolemy's Theorem)
Let . Let
be the center of the circle. Then
is twice the altitude of
. Since
is isosceles we can compute its area to be
, hence
.
Now by Ptolemy's Theorem we have This gives us:
Solution 6 (Trigonometry)
Since all three sides equal , they subtend three equal angles from the center. The right triangle between the center of the circle, a vertex, and the midpoint between two vertices has side lengths
by the Pythagorean Theorem. Thus, the sine of half of the subtended angle is
. Similarly, the cosine is
.
Since there are three sides, and since
,we seek to find
.
First,
and
by Pythagorean.
Solution 7 (Area)
For simplicity, scale everything down by a factor of 100. Let the inscribed trapezoid be , where
and
is the missing side length. Let
. If
and
are the midpoints of
and
, respectively, the height of the trapezoid is
. By the pythagorean theorem,
and
. Thus the height of the trapezoid is
, so the area is
. By Brahmagupta's formula, the area is
. Setting these two equal, we get
. Dividing both sides by
and then squaring, we get
. Expanding the right hand side and canceling the
terms gives us
. Rearranging and dividing by two, we get
. Squaring both sides, we get
. Rearranging, we get
. Dividing by 4 we get
. Factoring we get,
, and since
cannot be negative, we get
. Since
,
. Scaling up by 100, we get
.
Solution 8 (Cheap Solution - it doesn't make any cents!)
WLOG, let , and let ABCD be inscribed in a clrcle with radius
. We draw perpendiculars from
and
to
, and label the intersections
and
, respectively. We can see that
(because BCFE is a rectangle), and since
is clearly greater than 200, and and since
, which is part of segment
, is an integer, than we conclude that
is also an integer or of the form
. There is no reason for
to be of the form
because it seems too arbitrary. The only other integer choice is
.
See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 |
2016 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.