Difference between revisions of "2016 AMC 10A Problems/Problem 24"
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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)== | + | ==Solution 2 (Basic Algebra)== |
+ | <asy> | ||
+ | size(250); | ||
+ | defaultpen(linewidth(0.4)); | ||
+ | //Variable Declarations | ||
+ | real RADIUS; | ||
+ | pair A, B, C, D, O; | ||
+ | RADIUS=3; | ||
+ | |||
+ | //Variable Definitions | ||
+ | A=RADIUS*dir(148.414); | ||
+ | B=RADIUS*dir(109.471); | ||
+ | C=RADIUS*dir(70.529); | ||
+ | D=RADIUS*dir(31.586); | ||
+ | O=(0,0); | ||
+ | |||
+ | //Path Definitions | ||
+ | path quad= A -- B -- C -- D -- cycle; | ||
+ | |||
+ | //Initial Diagram | ||
+ | draw(Circle(O, RADIUS), linewidth(0.8)); | ||
+ | draw(quad, linewidth(0.8)); | ||
+ | label("$A$",A,W); | ||
+ | label("$B$",B,NW); | ||
+ | label("$C$",C,NE); | ||
+ | label("$D$",D,E); | ||
+ | label("$O$",O,S); | ||
+ | |||
+ | //Radii | ||
+ | draw(O--A); | ||
+ | draw(O--B); | ||
+ | draw(O--C); | ||
+ | draw(O--D); | ||
+ | |||
+ | </asy> | ||
+ | |||
+ | Let quadrilateral <math>ABCD</math> be inscribed in circle <math>O</math>, where <math>AD</math> is the side of unknown length. Draw the radii from center <math>O</math> to all four vertices of the quadrilateral, and draw the altitude of <math>\triangle BOC</math> such that it passes through side <math>AD</math> at the point <math>G</math> and meets side <math>BC</math> at the point <math>H</math>. | ||
+ | |||
+ | By the Pythagorean Theorem, the length of <math>OH</math> is | ||
+ | <cmath>\begin{align*} | ||
+ | \sqrt{CO^2 - HC^2} &= \sqrt{(200\sqrt{2})^2 - \left(\frac{200}{2}\right)^2} | ||
+ | \\ &= \sqrt{80000 - 10000} | ||
+ | \\ &= \sqrt{70000} | ||
+ | \\ &= 100\sqrt{7}. | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Note that <math>[ABCDO] = [AOB] + [BOC] + [COD] = [AOD] + [ABCD].</math> Let the length of <math>OG</math> be <math>h</math> and the length of <math>AD</math> be <math>x</math>; then we have that | ||
+ | |||
+ | <math> [AOB] + [BOC] + [COD] = \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2} = [AOD] + [ABCD].</math> | ||
+ | |||
+ | Furthermore, | ||
+ | <cmath>\begin{align*} | ||
+ | h &= \sqrt{OD^2 - GD^2} | ||
+ | \\ &= \sqrt{(200\sqrt{2})^2 - \left(\frac{x}{2}\right)^2} | ||
+ | \\ &= \sqrt{80000 - \frac{x^2}{4}} | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Substituting this value of <math>h</math> into the previous equation and evaluating for <math>x</math>, we get: | ||
+ | <cmath>\frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2}</cmath> | ||
+ | <cmath>\frac{3 \times 200 \times 100\sqrt{7}}{2} = \frac{x\sqrt{80000 - \frac{x^2}{4}}}{2} + \frac{\left(100\sqrt{7} - \sqrt{80000 - \frac{x^2}{4}}\right)(200 + x)}{2}</cmath> | ||
+ | <cmath>60000\sqrt{7} = \left(x\sqrt{80000 - \frac{x^2}{4}}\right) + \left(20000\sqrt{7}\right) + \left(100x\sqrt{7}\right) - \left(200\sqrt{80000 - \frac{x^2}{4}}\right) - \left(x\sqrt{80000 - \frac{x^2}{4}}\right)</cmath> | ||
+ | <cmath>40000\sqrt{7} = 100x\sqrt{7} - 200\sqrt{80000 - \frac{x^2}{4}}</cmath> | ||
+ | <cmath>400\sqrt{7} = x\sqrt{7} - 2\sqrt{80000 - \frac{x^2}{4}}</cmath> | ||
+ | <cmath>(x - 400)\sqrt{7} = 2\sqrt{80000 - \frac{x^2}{4}}</cmath> | ||
+ | <cmath>7(x-400)^2 = 4\left(80000 - \frac{x^2}{4}\right)</cmath> | ||
+ | <cmath>7x^2 - 5600x + 1120000 = 320000 - x^2</cmath> | ||
+ | <cmath>8x^2 - 5600x + 800000 = 0</cmath> | ||
+ | <cmath>x^2 - 700x + 100000 = 0</cmath> | ||
+ | |||
+ | The roots of this quadratic are found by using the quadratic formula: | ||
+ | <cmath>\begin{align*} | ||
+ | x &= \frac{-(-700) \pm \sqrt{(-700)^2 - 4 \times 1 \times 100000}}{2 \times 1} | ||
+ | \\ &= \frac{700 \pm \sqrt{490000 - 400000}}{2} | ||
+ | \\ &= \frac{700}{2} \pm \frac{\sqrt{90000}}{2} | ||
+ | \\ &= 350 \pm \frac{300}{2} | ||
+ | \\ &= 200, 500 | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | If the length of <math>AD</math> is <math>200</math>, then quadrilateral <math>ABCD</math> would be a square and thus, the radius of the circle would be | ||
+ | <cmath>\frac{\sqrt{200^2 + 200^2}}{2} = \frac{\sqrt{80000}}{2} = \frac{200\sqrt{2}}{2} = 100\sqrt{2}</cmath> | ||
+ | Which is a contradiction. Therefore, our answer is <math>\boxed{500}.</math> | ||
+ | |||
+ | ==Solution 3 (Trigonometry Bash)== | ||
<asy> | <asy> | ||
size(250); | size(250); | ||
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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 | + | ==Solution 4 (Easier trig)== |
<asy> | <asy> | ||
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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 | + | ==Solution 5 (Just Geometry)== |
<asy> | <asy> | ||
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<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 | + | ==Solution 6 (Ptolemy's Theorem)== |
<asy> | <asy> | ||
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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 | + | ==Solution 7 (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>. | ||
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<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 | + | ==Solution 8 (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 | + | ==Solution 9 (Cheap Solution - For when you are running out of time.)== |
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>. | ||
− | ==Solution | + | ==Solution 10 (Measuring)== |
A quick aside -- this problem takes about 90 seconds to solve with an accurate compass and a good ruler. Draw the diagram, to scale, and measure the last side. It becomes very clear that the answer is <math>\boxed{\textbf{(E)}\text{ 500}}</math>. | A quick aside -- this problem takes about 90 seconds to solve with an accurate compass and a good ruler. Draw the diagram, to scale, and measure the last side. It becomes very clear that the answer is <math>\boxed{\textbf{(E)}\text{ 500}}</math>. |
Revision as of 14:03, 12 February 2017
Contents
- 1 Problem
- 2 Solution 1 (Algebra)
- 3 Solution 2 (Basic Algebra)
- 4 Solution 3 (Trigonometry Bash)
- 5 Solution 4 (Easier trig)
- 6 Solution 5 (Just Geometry)
- 7 Solution 6 (Ptolemy's Theorem)
- 8 Solution 7 (Trigonometry)
- 9 Solution 8 (Area)
- 10 Solution 9 (Cheap Solution - For when you are running out of time.)
- 11 Solution 10 (Measuring)
- 12 See Also
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 (Basic Algebra)
Let quadrilateral be inscribed in circle
, where
is the side of unknown length. Draw the radii from center
to all four vertices of the quadrilateral, and draw the altitude of
such that it passes through side
at the point
and meets side
at the point
.
By the Pythagorean Theorem, the length of is
Note that Let the length of
be
and the length of
be
; then we have that
Furthermore,
Substituting this value of into the previous equation and evaluating for
, we get:
The roots of this quadratic are found by using the quadratic formula:
If the length of is
, then quadrilateral
would be a square and thus, the radius of the circle would be
Which is a contradiction. Therefore, our answer is
Solution 3 (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 4 (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 5 (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 6 (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 7 (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 8 (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 9 (Cheap Solution - For when you are running out of time.)
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
.
Solution 10 (Measuring)
A quick aside -- this problem takes about 90 seconds to solve with an accurate compass and a good ruler. Draw the diagram, to scale, and measure the last side. It becomes very clear that the answer is .
-- Jonathan Ko
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.