Difference between revisions of "2016 AMC 10A Problems/Problem 24"
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+ | {{duplicate|[[2016 AMC 10A Problems#Problem 24|2016 AMC 10A #24]] and [[2016 AMC 12A Problems#Problem 21|2016 AMC 12A #21]]}} | ||
+ | |||
==Problem== | ==Problem== | ||
A quadrilateral is inscribed in a circle of radius <math>200\sqrt{2}</math>. Three of the sides of this quadrilateral have length <math>200</math>. What is the length of the fourth side? | A quadrilateral is inscribed in a circle of radius <math>200\sqrt{2}</math>. Three of the sides of this quadrilateral have length <math>200</math>. What is the length of the fourth side? | ||
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<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)== | + | ==Solution 1 == |
+ | |||
+ | <asy> | ||
+ | size(250); | ||
+ | defaultpen(linewidth(0.4)); | ||
+ | //Variable Declarations | ||
+ | real RADIUS; | ||
+ | pair A, B, C, D, E, F, 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,ENE); | ||
+ | label("$O$",O,S); | ||
+ | label("$\theta$",O,3N); | ||
+ | |||
+ | //Radii | ||
+ | draw(O--A); | ||
+ | draw(O--B); | ||
+ | draw(O--C); | ||
+ | draw(O--D); | ||
+ | |||
+ | //Construction | ||
+ | E=extension(B,O,A,D); | ||
+ | |||
+ | label("$E$",E,NE); | ||
+ | |||
+ | F=extension(C,O,A,D); | ||
+ | |||
+ | label("$F$",F,NE); | ||
+ | |||
+ | |||
+ | //Angle marks | ||
+ | draw(anglemark(C,O,B)); | ||
+ | |||
+ | </asy> | ||
+ | |||
+ | Let <math>AD</math> intersect <math>OB</math> at <math>E</math> and <math>OC</math> at <math>F.</math> | ||
+ | |||
+ | |||
+ | <math>\overarc{AB}= \overarc{BC}= \overarc{CD}=\theta</math> | ||
+ | |||
+ | <math>\angle{BAD}=\frac{1}{2} \cdot \overarc{BCD}=\theta=\angle{AOB}</math> | ||
+ | |||
+ | |||
+ | From there, <math>\triangle{OAB} \sim \triangle{ABE}</math>, thus: | ||
+ | |||
+ | <math>\frac{OA}{AB} = \frac{AB}{BE} = \frac{OB}{AE}</math> | ||
+ | |||
+ | <math>OA = OB</math> because they are both radii of <math>\odot{O}</math>. Since <math>\frac{OA}{AB} = \frac{OB}{AE}</math>, we have that <math>AB = AE</math>. Similarly, <math>CD = DF</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 2 (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. | ||
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Construct quadrilateral <math>ABCD</math> on the circle with <math>AD</math> 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 <math>O</math> to <math>A,B,C,</math> and <math>D</math>. Let the intersection of <math>BD</math> and <math>OC</math> be point <math>E</math>. Notice that <math>BD</math> and <math>OC</math> are perpendicular because <math>BCDO</math> is a kite. | Construct quadrilateral <math>ABCD</math> on the circle with <math>AD</math> 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 <math>O</math> to <math>A,B,C,</math> and <math>D</math>. Let the intersection of <math>BD</math> and <math>OC</math> be point <math>E</math>. Notice that <math>BD</math> and <math>OC</math> are perpendicular because <math>BCDO</math> is a kite. | ||
− | We set lengths <math>BE=ED</math> equal to <math>x</math>. By the Pythagorean Theorem, | + | We set lengths <math>BE=ED</math> equal to <math>x</math> (Solution 1.1 begins from here). By the Pythagorean Theorem, |
<cmath>\sqrt{1^2-x^2}+\sqrt{(\sqrt{2})^2-x^2}=\sqrt{2}</cmath> | <cmath>\sqrt{1^2-x^2}+\sqrt{(\sqrt{2})^2-x^2}=\sqrt{2}</cmath> | ||
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<cmath>x=\frac{\sqrt{14}}{4}</cmath> | <cmath>x=\frac{\sqrt{14}}{4}</cmath> | ||
− | By Ptolemy's Theorem, | + | By [[Ptolemy's Theorem]], |
<cmath>AB \cdot CD + BC \cdot AD = AC \cdot BD = BD^2 = (2 \cdot BE)^2</cmath> | <cmath>AB \cdot CD + BC \cdot AD = AC \cdot BD = BD^2 = (2 \cdot BE)^2</cmath> | ||
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<cmath>AD=\frac{5}{2}</cmath> | <cmath>AD=\frac{5}{2}</cmath> | ||
− | Finally, we multiply back the <math>200</math> that we divided by at the beginning of the problem to get <math>AD=\boxed{ | + | Finally, we multiply back the <math>200</math> that we divided by at the beginning of the problem to get <math>AD=\boxed{\textbf{(E) } 500}</math>. |
− | ==Solution | + | ==Solution 3 (HARD Algebra)== |
<asy> | <asy> | ||
size(250); | size(250); | ||
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</asy> | </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 | + | 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 point <math>H</math>. |
By the Pythagorean Theorem, the length of <math>OH</math> is | By the Pythagorean Theorem, the length of <math>OH</math> is | ||
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\end{align*}</cmath> | \end{align*}</cmath> | ||
− | If the length of <math>AD</math> is <math>200</math>, then | + | If the length of <math>AD</math> is <math>200</math>, then <math>ABCD</math> would be a square. 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> | <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> | Which is a contradiction. Therefore, our answer is <math>\boxed{500}.</math> | ||
− | ==Solution | + | ==Solution 4 (Trigonometry Bash)== |
<asy> | <asy> | ||
size(250); | size(250); | ||
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</asy> | </asy> | ||
− | Construct quadrilateral <math>ABCD</math> on the circle with <math>AD</math> 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 <math>O</math> to <math>A,B,C,</math> and <math>D</math>. Apply law of cosines on <math>\Delta BOC</math>; let <math> \theta = \angle BOC</math>. We get the following equation: <cmath>(BC)^{2}=(OB)^{2}+(OC)^{2}-2\cdot OB \cdot OC\cdot \cos\theta</cmath> Substituting the values in, we get <cmath>(200)^{2}=2\cdot (200)^{2}+ 2\cdot (200)^{2}- 2\cdot 2\cdot (200)^{2}\cdot \cos\theta</cmath> Canceling out, we get <cmath>\cos\theta=\frac{3}{4}</cmath> | + | Construct quadrilateral <math>ABCD</math> on the circle with <math>AD</math> 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 <math>O</math> to <math>A,B,C,</math> and <math>D</math>. Apply the [[law of cosines]] on <math>\Delta BOC</math>; let <math> \theta = \angle BOC</math>. We get the following equation: <cmath>(BC)^{2}=(OB)^{2}+(OC)^{2}-2\cdot OB \cdot OC\cdot \cos\theta</cmath> Substituting the values in, we get <cmath>(200)^{2}=2\cdot (200)^{2}+ 2\cdot (200)^{2}- 2\cdot 2\cdot (200)^{2}\cdot \cos\theta</cmath> Canceling out, we get <cmath>\cos\theta=\frac{3}{4}</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- (2\sin\theta \cos\theta) \cdot \sin \theta</cmath> <cmath>\cos 3\theta= \cos\theta (\cos 2\theta-2\sin^{2}\theta)=\cos\theta (2\cos^{2}\theta-3+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- (2\sin\theta \cos\theta) \cdot \sin \theta</cmath> <cmath>\cos 3\theta= \cos\theta (\cos 2\theta-2\sin^{2}\theta)=\cos\theta (2\cos^{2}\theta-3+2\cos^{2}\theta)</cmath> <cmath>\Rightarrow \cos 3\theta=4\cos^{3}\theta-3\cos\theta</cmath> It is useful to memorize the triple angle formulas (<math>\cos 3\theta=4\cos^{3}\theta-3\cos\theta, \sin 3\theta=3\sin\theta-4\sin^{3}\theta</math>). 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 5 (Easier Trigonometry)== |
<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 6 (Ptolemy's Theorem)== |
+ | <asy> | ||
+ | pathpen = black; pointpen = black; | ||
+ | size(6cm); | ||
+ | draw(unitcircle); | ||
+ | pair A = D("A", dir(50), dir(50)); | ||
+ | pair B = D("B", dir(90), dir(90)); | ||
+ | pair C = D("C", dir(130), dir(130)); | ||
+ | pair D = D("D", dir(170), dir(170)); | ||
+ | pair O = D("O", (0,0), dir(-90)); | ||
+ | draw(A--C, red); | ||
+ | draw(B--D, blue+dashed); | ||
+ | draw(A--B--C--D--cycle); | ||
+ | draw(A--O--C); | ||
+ | draw(O--B); | ||
+ | </asy> | ||
+ | |||
+ | Let <math>s = 200</math>. Let <math>O</math> be the center of the circle. Then <math>AC</math> is twice the altitude of <math>\triangle OBC</math> to <math>\overline{OB}</math>. Since <math>\triangle OBC</math> is isosceles we can compute its area to be <math>\frac{s^2 \sqrt{7}}{4}</math>, hence <math>CA = 2 \cdot \tfrac{2 \cdot s^2\sqrt7/4}{s\sqrt2} = s\sqrt{\frac{7}{2}}</math>. | ||
+ | |||
+ | Now by Ptolemy's Theorem we have <math>CA^2 = s^2 + AD \cdot s \implies AD = \left(\frac{7}{2}-1\right)s.</math> This gives us: <cmath>\boxed{\textbf{(E) } 500.}</cmath> | ||
+ | |||
+ | ==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 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>. | ||
+ | First, <math>\sin 2\theta=2\sin\theta\cos\theta=2\cdot\left(\frac{\sqrt{2}}{4}\right)\left(\frac{\sqrt{14}}{4}\right)=\frac{2\sqrt{2}\sqrt{14}}{16}=\frac{\sqrt{7}}{4}</math> and <math>\cos 2\theta=\frac{3}{4}</math> by Pythagorean. | ||
+ | <cmath>\sin 3\theta=\sin(2\theta+\theta)=\sin 2\theta\cos\theta+\sin \theta\cos 2\theta=\frac{\sqrt{7}}{4}\left(\frac{\sqrt{14}}{4}\right)+\frac{\sqrt{2}}{4}\left(\frac{3}{4}\right)=\frac{7\sqrt{2}+3\sqrt{2}}{16}=\frac{5\sqrt{2}}{8}</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 8 (Area By Brahmagupta's Formula)== | ||
+ | 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 9 (Similar Triangles)== | ||
+ | <asy> | ||
+ | size(250); | ||
+ | defaultpen(linewidth(0.4)); | ||
+ | //Variable Declarations, L is used to write alpha= statement | ||
+ | real RADIUS; | ||
+ | pair A, B, C, D, E, F, O, L; | ||
+ | RADIUS=3; | ||
+ | |||
+ | //Variable Definitions | ||
+ | A=RADIUS*dir(148.414); | ||
+ | B=RADIUS*dir(109.471); | ||
+ | C=RADIUS*dir(70.529); | ||
+ | D=RADIUS*dir(31.586); | ||
+ | E=extension(A,D,O,B); | ||
+ | F=extension(A,D,O,C); | ||
+ | L=midpoint(C--D); | ||
+ | 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,NW); | ||
+ | label("$B$",B,NW); | ||
+ | label("$C$",C,NE); | ||
+ | label("$D$",D,NE); | ||
+ | label("$E$",E,SW); | ||
+ | label("$F$",F,SE); | ||
+ | label("$O$",O,SE); | ||
+ | dot(O,linewidth(5)); | ||
+ | |||
+ | //Radii | ||
+ | draw(O--A); | ||
+ | draw(O--B); | ||
+ | draw(O--C); | ||
+ | draw(O--D); | ||
+ | |||
+ | //Construction | ||
+ | label("$\alpha = 90-\frac{\theta}{2}$",L,5NE,rgb(128, 0, 0)); | ||
+ | draw(anglemark(C,O,B)); | ||
+ | label("$\theta$",O,3N); | ||
+ | draw(anglemark(E,F,O)); | ||
+ | label("$\alpha$",F,3SW); | ||
+ | draw(anglemark(D,F,C)); | ||
+ | label("$\alpha$",F,3NE); | ||
+ | draw(anglemark(F,C,D)); | ||
+ | label("$\alpha$",C,3SSE); | ||
+ | draw(anglemark(C,D,F)); | ||
+ | label("$\theta$",(RADIUS-0.04)*dir(31.586),3WNW); | ||
+ | </asy> | ||
+ | Label the points as shown, and let <math>\angle{EOF} = \theta</math>. Since <math>\overline{OB} = \overline{OC}</math>, and <math>\triangle{OFE} \sim \triangle{OCB}</math>, we get that <math>\angle{EFO} = 90-\frac{\theta}{2}</math>. We assign <math>\alpha</math> to <math>90-\frac{\theta}{2}</math> for simplicity. | ||
+ | From here, by vertical angles <math>\angle{CFD} = \alpha</math>. Also, since <math>\triangle{OCB} \cong \triangle{ODC}</math>, <math>\angle{OCD} = \alpha</math>. This means that <math>\angle{CDF} = 180-2\alpha = \theta</math>, which leads to <math>\triangle{OCB} \sim \triangle{DCF}</math>. | ||
+ | Since we know that <math>\overline{CD} = 200</math>, <math>\overline{DF} = 200</math>, and by similar reasoning <math>\overline{AE} = 200</math>. | ||
+ | Finally, again using similar triangles, we get that <math>\overline{CF} = 100\sqrt{2}</math>, which means that <math>\overline{OF} = \overline{OC} - \overline{CF} = 200\sqrt{2} - 100\sqrt{2} = 100\sqrt{2}</math>. We can again apply similar triangles (or use Power of a Point) to get <math>\overline{EF} = 100</math>, and finally <math>\overline{AD} = \overline{AE}+\overline{EF}+\overline{FD} = 200+100+200=\boxed{\textbf{(E)}500}</math> - ColtsFan10 | ||
+ | |||
+ | ==Solution 10 (Complex Numbers)== | ||
+ | |||
+ | We first scale down by a factor of <math>200\sqrt{2}</math>. Let the vertices of the quadrilateral be <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>, so that <math>AD</math> is the length of the fourth side. We draw this in the complex plane so that <math>D</math> corresponds to the complex number <math>1</math>, and we let <math>C</math> correspond to the complex number <math>z</math>. Then, <math>A</math> corresponds to <math>z^3</math> and <math>B</math> corresponds to <math>z^2</math>. We are given that <math>\lvert z \rvert = 1</math> and <math>\lvert z-1 \rvert = 1/\sqrt{2}</math>, and we wish to find <math>\lvert z^3 - 1 \rvert=\lvert z^2+z+1\rvert \cdot \lvert z-1 \rvert=\lvert (z^2+z+1)/\sqrt{2} \rvert</math>. Let <math>z=a+bi</math>, where <math>a</math> and <math>b</math> are real numbers. Then, <math>a^2+b^2=1</math> and <math>a^2-2a+1+b^2=1/2</math>; solving for <math>a</math> and <math>b</math> yields <math>a=3/4</math> and <math>b=\sqrt{7}/4</math>. Thus, <math>AD = \lvert z^3 - 1 \rvert = \lvert (z^2+z+1)/\sqrt{2} \rvert = \lvert (15/8 + 5\sqrt{7}/8 \cdot i)/\sqrt{2} \rvert = \frac{5\sqrt{2}}{4}</math>. Scaling back up gives us a final answer of <math>\frac{5\sqrt{2}}{4} \cdot 200\sqrt{2} = \boxed{\textbf{(E)} 500}</math>. | ||
+ | |||
+ | ~ Leo.Euler | ||
+ | |||
+ | ==Solution 11 (Trignometry + Ptolemy’s)== | ||
+ | |||
+ | Let angle <math>C</math> be <math>2a</math>. This way <math>BD</math> will be <math>400sin(a)</math>. Now we can trig bash. As the circumradius of triangle <math>BCD</math> is <math>200\sqrt{2}</math>, we can use the formula <cmath>R=\frac{abc}{4A}</cmath> and <cmath>A=\frac{absin(C)}{2}</cmath> and plug in all the values we got to get <cmath>200\sqrt{2}=\frac{200^2 \cdot 400sin(a)}{4 \cdot (\frac{200^2 sin(2a)}{2})}</cmath>. This boils down to <cmath>\sqrt{2}=\frac{sin(a)}{sin{2a}}</cmath>. This expression can further be simplified by the trig identity <cmath>sin(2a)=2sin(a)cos(a)</cmath>. This leads to the final simplified form <cmath>2\sqrt{2}=\frac{1}{cos(a)}</cmath>. Solving this expression gives us <cmath>cos(a)=\frac{\sqrt{2}}{4}</cmath>. However, as we want <math>sin(a)</math>, we use the identity <math>sin^2+cos^2=1</math>, and substitute to get that <math>sin(a)=\frac{\sqrt{14}}{4}</math>, and therefore BD is <math>100\sqrt{14}</math>. | ||
+ | |||
+ | Then, as <math>ABCD</math> is a cyclic quadrilateral, we can use [[Ptolemy’s Theorem]] (with <math>AD=x</math>) to get <cmath>14 \cdot 100^2=200x+200^2</cmath>. Finally, we solve to get <math>\boxed{\textbf{(E) } 500}</math>. | ||
+ | |||
+ | -dragoon | ||
+ | |||
+ | ==Solution 12 (Simple Trigonometry with Geometric Observations)== | ||
<asy> | <asy> | ||
size(250); | size(250); | ||
Line 260: | Line 432: | ||
//Variable Declarations | //Variable Declarations | ||
real RADIUS; | real RADIUS; | ||
− | pair A, B, C, D | + | pair A, B, C, D, O; |
RADIUS=3; | RADIUS=3; | ||
Line 279: | Line 451: | ||
label("$B$",B,NW); | label("$B$",B,NW); | ||
label("$C$",C,NE); | label("$C$",C,NE); | ||
− | label("$D$",D, | + | label("$D$",D,E); |
label("$O$",O,S); | label("$O$",O,S); | ||
− | |||
//Radii | //Radii | ||
Line 289: | Line 460: | ||
draw(O--D); | draw(O--D); | ||
− | // | + | </asy> |
− | + | Claim: <math>[ABCD]</math> is an isosceles trapezoid. | |
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
+ | Proof: Notice that <math>[ABCD]</math> is cyclic, triangle <math>BOC</math> is isosceles, and triangle <math>AOB</math> is congruent to <math>DOC</math> by SSS congruence. Therefore, <math>\angle BAD = 180 - \angle BCD = 180-(\angle BCO + \angle DCO)=180-(\angle CBO+\angle ABO) = 180 - \angle ABC = \angle CDA</math>. Hence, <math>[ABCD]</math> is an isosceles trapezoid. | ||
− | // | + | Let <math>\angle CDA=\alpha</math>. Notice that the length of the altitude from <math>C</math> to <math>AD</math> is <math>200sin(\alpha)</math>. Furthermore, the length of the altitude from <math>O</math> to <math>BC</math> is <math>100\sqrt{7}</math> by the Pythagorean theorem. Therefore, the length of the altitude from <math>O</math> to <math>AD</math> is <math>100\sqrt{7}-200sin\alpha</math>. Let <math>F</math> the feet of the altitude from <math>O</math> to <math>AD</math>. Then, <math>FD=(200+400cos(\alpha))/2=100+200cos(\alpha)</math>, because <math>AOD</math> is isosceles. |
− | |||
− | </ | + | Therefore, by the Pythagorean theorem, <math>(100+200cos(\alpha))^2+(100\sqrt{7}-200sin(\alpha))^2=80000</math>. Simplifying, we have <math>1+cos(\alpha)=sin(\alpha) \cdot sqrt{7} \implies cos^2(\alpha)+2cos(\alpha)+1=sin^2(\alpha) \cdot 7 = 7-7cos^2(\alpha) \implies 8cos^2(\alpha)+2cos(\alpha) - 6 =0</math>. Solving this quadratic, we have <math>cos(\alpha)=\frac{3}{4}, -1</math>, but <math>0<\alpha<180 \implies cos(\alpha)=3/4</math>. Therefore, <math>AD=200cos(\alpha)+200cos(\alpha)+200=\boxed{500}</math> |
− | + | - [mathMagicOPS] | |
+ | ==Remark (Morley's Trisector Theorem)== | ||
− | + | This problem is related to [http://www.cut-the-knot.org/triangle/Morley/Naraniengar.shtml M. T. Naraniengar's proof] of [https://en.wikipedia.org/wiki/Morley%27s_trisector_theorem Morley's Trisector Theorem]. This problem is taken from the figure of the Lemma of M. T. Naraniengar's proof, as shown below. | |
− | + | [[File:NaraniengarLemma.gif|200px]] | |
+ | If four points <math>Y'</math>, <math>Z</math>, <math>Y</math>, <math>Z'</math> satisfy the conditions | ||
− | + | <math>\quad</math> <math>1.</math> <math>Y'Z = ZY = YZ'</math> and | |
− | <math>\ | + | <math>\quad</math> <math>2.</math> <math>\angle YZY'</math> = <math>\angle Z'YZ</math> = <math>180^{\circ} - 2a > 60^{\circ}</math> |
− | + | then they lie on a circle. | |
− | |||
− | + | The Lemma is used to prove Morley's Trisector Theorem by constructing an equilateral triangle at <math>YZ</math> and extending <math>AY'</math> and <math>AZ'</math> as shown below. | |
− | + | [[File:NaraniengarTheorem.gif|400px]] | |
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | + | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | |
− | + | ==Video Solution by AoPS (Deven Ware)== | |
+ | https://www.youtube.com/watch?v=hpSyHZwsteM | ||
− | + | ==Video Solution by Walt S.== | |
+ | https://www.youtube.com/watch?v=3iDqR9YNNkU | ||
− | + | == Video Solution (Ptolemy’s Theorem) == | |
+ | https://youtu.be/NsQbhYfGh1Q?t=5094 | ||
− | ==Solution | + | ~ pi_is_3.14 |
− | + | ==Video Solution by TheBeautyofMath== | |
− | + | https://youtu.be/gCmQlaiEG5A | |
− | |||
− | |||
− | |||
− | ==Solution | + | ~IceMatrix |
− | + | ==Video Solution by Punxsutawney Phil== | |
+ | https://www.youtube.com/watch?v=st6HIgDWgX4 | ||
− | ==Solution | + | == Video Solution by OmegaLearn == |
+ | https://youtu.be/NsQbhYfGh1Q?t=5094 | ||
− | + | ~ pi_is_3.14 | |
==See Also== | ==See Also== | ||
+ | |||
{{AMC10 box|year=2016|ab=A|num-b=23|num-a=25}} | {{AMC10 box|year=2016|ab=A|num-b=23|num-a=25}} | ||
{{AMC12 box|year=2016|ab=A|num-b=20|num-a=22}} | {{AMC12 box|year=2016|ab=A|num-b=20|num-a=22}} | ||
+ | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:17, 26 April 2024
- The following problem is from both the 2016 AMC 10A #24 and 2016 AMC 12A #21, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Algebra)
- 4 Solution 3 (HARD Algebra)
- 5 Solution 4 (Trigonometry Bash)
- 6 Solution 5 (Easier Trigonometry)
- 7 Solution 6 (Ptolemy's Theorem)
- 8 Solution 7 (Trigonometry)
- 9 Solution 8 (Area By Brahmagupta's Formula)
- 10 Solution 9 (Similar Triangles)
- 11 Solution 10 (Complex Numbers)
- 12 Solution 11 (Trignometry + Ptolemy’s)
- 13 Solution 12 (Simple Trigonometry with Geometric Observations)
- 14 Remark (Morley's Trisector Theorem)
- 15 Video Solution by AoPS (Deven Ware)
- 16 Video Solution by Walt S.
- 17 Video Solution (Ptolemy’s Theorem)
- 18 Video Solution by TheBeautyofMath
- 19 Video Solution by Punxsutawney Phil
- 20 Video Solution by OmegaLearn
- 21 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
Let intersect
at
and
at
From there, , thus:
because they are both radii of
. Since
, we have that
. Similarly,
.
and
, so
Solution 2 (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
(Solution 1.1 begins from here). By the Pythagorean Theorem,
We solve for :
Substituting values,
Finally, we multiply back the that we divided by at the beginning of the problem to get
.
Solution 3 (HARD 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 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
would be a square. Thus, the radius of the circle would be
Which is a contradiction. Therefore, our answer is
Solution 4 (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 the 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
:
It is useful to memorize the triple angle formulas (
). Plugging in
, we get
. Now, applying law of cosines on triangle
, we get
Solution 5 (Easier Trigonometry)
Construct quadrilateral on the circle
with
being the desired side. Then, drop perpendiculars from
and
to the extended line of
and let these points be
and
, respectively. Also, let
. From the Law of Cosines on
, we have
.
Now, since is isosceles with
, we have that
. In addition, we know that
as they are both equal to
and
as they are both radii of the same circle. By SSS Congruence, we have that
, so we have that
, so
.
Thus, we have , so
. Similarly,
, and
.
Solution 6 (Ptolemy's Theorem)
Let . Let
be the center of the circle. Then
is twice the altitude of
to
. 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 By Brahmagupta's Formula)
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 (Similar Triangles)
Label the points as shown, and let
. Since
, and
, we get that
. We assign
to
for simplicity.
From here, by vertical angles
. Also, since
,
. This means that
, which leads to
.
Since we know that
,
, and by similar reasoning
.
Finally, again using similar triangles, we get that
, which means that
. We can again apply similar triangles (or use Power of a Point) to get
, and finally
- ColtsFan10
Solution 10 (Complex Numbers)
We first scale down by a factor of . Let the vertices of the quadrilateral be
,
,
, and
, so that
is the length of the fourth side. We draw this in the complex plane so that
corresponds to the complex number
, and we let
correspond to the complex number
. Then,
corresponds to
and
corresponds to
. We are given that
and
, and we wish to find
. Let
, where
and
are real numbers. Then,
and
; solving for
and
yields
and
. Thus,
. Scaling back up gives us a final answer of
.
~ Leo.Euler
Solution 11 (Trignometry + Ptolemy’s)
Let angle be
. This way
will be
. Now we can trig bash. As the circumradius of triangle
is
, we can use the formula
and
and plug in all the values we got to get
. This boils down to
. This expression can further be simplified by the trig identity
. This leads to the final simplified form
. Solving this expression gives us
. However, as we want
, we use the identity
, and substitute to get that
, and therefore BD is
.
Then, as is a cyclic quadrilateral, we can use Ptolemy’s Theorem (with
) to get
. Finally, we solve to get
.
-dragoon
Solution 12 (Simple Trigonometry with Geometric Observations)
Claim:
is an isosceles trapezoid.
Proof: Notice that is cyclic, triangle
is isosceles, and triangle
is congruent to
by SSS congruence. Therefore,
. Hence,
is an isosceles trapezoid.
Let . Notice that the length of the altitude from
to
is
. Furthermore, the length of the altitude from
to
is
by the Pythagorean theorem. Therefore, the length of the altitude from
to
is
. Let
the feet of the altitude from
to
. Then,
, because
is isosceles.
Therefore, by the Pythagorean theorem, . Simplifying, we have
. Solving this quadratic, we have
, but
. Therefore,
- [mathMagicOPS]
Remark (Morley's Trisector Theorem)
This problem is related to M. T. Naraniengar's proof of Morley's Trisector Theorem. This problem is taken from the figure of the Lemma of M. T. Naraniengar's proof, as shown below.
If four points ,
,
,
satisfy the conditions
and
=
=
then they lie on a circle.
The Lemma is used to prove Morley's Trisector Theorem by constructing an equilateral triangle at and extending
and
as shown below.
Video Solution by AoPS (Deven Ware)
https://www.youtube.com/watch?v=hpSyHZwsteM
Video Solution by Walt S.
https://www.youtube.com/watch?v=3iDqR9YNNkU
Video Solution (Ptolemy’s Theorem)
https://youtu.be/NsQbhYfGh1Q?t=5094
~ pi_is_3.14
Video Solution by TheBeautyofMath
~IceMatrix
Video Solution by Punxsutawney Phil
https://www.youtube.com/watch?v=st6HIgDWgX4
Video Solution by OmegaLearn
https://youtu.be/NsQbhYfGh1Q?t=5094
~ pi_is_3.14
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.