Difference between revisions of "2016 AIME I Problems/Problem 9"
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− | ==Problem== | + | ==Problem 9 (WARNING: CORRUPTED PROBLEM)== |
Triangle <math>ABC</math> has <math>AB=40,AC=31,</math> and <math>\sin{A}=\frac{1}{5}</math>. This triangle is inscribed in rectangle <math>AQRS</math> with <math>B</math> on <math>\overline{QR}</math> and <math>C</math> on <math>\overline{RS}</math>. Find the maximum possible area of <math>AQRS</math>. | Triangle <math>ABC</math> has <math>AB=40,AC=31,</math> and <math>\sin{A}=\frac{1}{5}</math>. This triangle is inscribed in rectangle <math>AQRS</math> with <math>B</math> on <math>\overline{QR}</math> and <math>C</math> on <math>\overline{RS}</math>. Find the maximum possible area of <math>AQRS</math>. | ||
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− | + | ==Solution 1== | |
Note that if angle <math>BAC</math> is obtuse, it would be impossible for the triangle to inscribed in a rectangle. This can easily be shown by drawing triangle ABC, where <math>A</math> is obtuse. Therefore, angle A is acute. Let angle <math>CAS=n</math> and angle <math>BAQ=m</math>. Then, <math>\overline{AS}=31\cos(n)</math> and <math>\overline{AQ}=40\cos(m)</math>. Then the area of rectangle <math>AQRS</math> is <math>1240\cos(m)\cos(n)</math>. By product-to-sum, <math>\cos(m)\cos(n)=\frac{1}{2}(\cos(m+n)+\cos(m-n))</math>. <math>\cos(m+n)=\sin(90-m-n)=\sin(BAC)=\frac{1}{5}</math>. The maximum possible value of <math>\cos(m-n)</math> is 1, which occurs when <math>m=n</math>. Thus the maximum possible value of <math>\cos(m)\cos(n)</math> is <math>\frac{1}{2}(\frac{1}{5}+1)=\frac{3}{5}</math> so the maximum possible area of <math>AQRS</math> is <math>1240\times{\frac{3}{5}}=\fbox{744}</math>. | Note that if angle <math>BAC</math> is obtuse, it would be impossible for the triangle to inscribed in a rectangle. This can easily be shown by drawing triangle ABC, where <math>A</math> is obtuse. Therefore, angle A is acute. Let angle <math>CAS=n</math> and angle <math>BAQ=m</math>. Then, <math>\overline{AS}=31\cos(n)</math> and <math>\overline{AQ}=40\cos(m)</math>. Then the area of rectangle <math>AQRS</math> is <math>1240\cos(m)\cos(n)</math>. By product-to-sum, <math>\cos(m)\cos(n)=\frac{1}{2}(\cos(m+n)+\cos(m-n))</math>. <math>\cos(m+n)=\sin(90-m-n)=\sin(BAC)=\frac{1}{5}</math>. The maximum possible value of <math>\cos(m-n)</math> is 1, which occurs when <math>m=n</math>. Thus the maximum possible value of <math>\cos(m)\cos(n)</math> is <math>\frac{1}{2}(\frac{1}{5}+1)=\frac{3}{5}</math> so the maximum possible area of <math>AQRS</math> is <math>1240\times{\frac{3}{5}}=\fbox{744}</math>. | ||
− | + | ==Solution 2== | |
As above, we note that angle <math>A</math> must be acute. Therefore, let <math>A</math> be the origin, and suppose that <math>Q</math> is on the positive <math>x</math> axis and <math>S</math> is on the positive <math>y</math> axis. We approach this using complex numbers. Let <math>w=\text{cis} A</math>, and let <math>z</math> be a complex number with <math>|z|=1</math>, <math>\text{Arg}(z)\ge 0^\circ</math> and <math>\text{Arg}(zw)\le90^\circ</math>. Then we represent <math>B</math> by <math>40z</math> and <math>C</math> by <math>31zw</math>. The coordinates of <math>Q</math> and <math>S</math> depend on the real part of <math>40z</math> and the imaginary part of <math>31zw</math>. Thus | As above, we note that angle <math>A</math> must be acute. Therefore, let <math>A</math> be the origin, and suppose that <math>Q</math> is on the positive <math>x</math> axis and <math>S</math> is on the positive <math>y</math> axis. We approach this using complex numbers. Let <math>w=\text{cis} A</math>, and let <math>z</math> be a complex number with <math>|z|=1</math>, <math>\text{Arg}(z)\ge 0^\circ</math> and <math>\text{Arg}(zw)\le90^\circ</math>. Then we represent <math>B</math> by <math>40z</math> and <math>C</math> by <math>31zw</math>. The coordinates of <math>Q</math> and <math>S</math> depend on the real part of <math>40z</math> and the imaginary part of <math>31zw</math>. Thus | ||
<cmath>[AQRS]=\Re(40z)\cdot \Im(31zw)=1240\left(\frac{z+\overline{z}}{2}\right)\left(\frac{zw-\overline{zw}}{2i}\right).</cmath> | <cmath>[AQRS]=\Re(40z)\cdot \Im(31zw)=1240\left(\frac{z+\overline{z}}{2}\right)\left(\frac{zw-\overline{zw}}{2i}\right).</cmath> | ||
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Now as <math>w=\text{cis}A</math>, we know that <math>\Im(w)=\frac15</math>. Also, <math>|z^2w|=1</math>, so the maximum possible imaginary part of <math>z^2w</math> is <math>1</math>. This is clearly achievable under our conditions on <math>z</math>. Therefore, the maximum possible area of <math>AQRS</math> is <math>620(1+\tfrac15)=\boxed{744}</math>. | Now as <math>w=\text{cis}A</math>, we know that <math>\Im(w)=\frac15</math>. Also, <math>|z^2w|=1</math>, so the maximum possible imaginary part of <math>z^2w</math> is <math>1</math>. This is clearly achievable under our conditions on <math>z</math>. Therefore, the maximum possible area of <math>AQRS</math> is <math>620(1+\tfrac15)=\boxed{744}</math>. | ||
− | + | ==Solution 3 (With Calculus)== | |
Let <math>\theta</math> be the angle <math>\angle BAQ</math>. The height of the rectangle then can be expressed as <math>h = 31 \sin (A+\theta)</math>, and the length of the rectangle can be expressed as <math>l = 40\cos \theta</math>. The area of the rectangle can then be written as a function of <math>\theta</math>, <math>[AQRS] = a(\theta) = 31\sin (A+\theta)\cdot 40 \cos \theta = 1240 \sin (A+\theta) \cos \theta</math>. For now, we will ignore the <math>1240</math> and focus on the function <math>f(\theta) = \sin (A+\theta) \cos \theta = (\sin A \cos \theta + \cos A \sin \theta)(\cos \theta) = \sin A \cos^2 \theta + \cos A \sin \theta \cos \theta = \sin A \cos^2 \theta + \frac{1}{2} \cos A \sin 2\theta</math>. | Let <math>\theta</math> be the angle <math>\angle BAQ</math>. The height of the rectangle then can be expressed as <math>h = 31 \sin (A+\theta)</math>, and the length of the rectangle can be expressed as <math>l = 40\cos \theta</math>. The area of the rectangle can then be written as a function of <math>\theta</math>, <math>[AQRS] = a(\theta) = 31\sin (A+\theta)\cdot 40 \cos \theta = 1240 \sin (A+\theta) \cos \theta</math>. For now, we will ignore the <math>1240</math> and focus on the function <math>f(\theta) = \sin (A+\theta) \cos \theta = (\sin A \cos \theta + \cos A \sin \theta)(\cos \theta) = \sin A \cos^2 \theta + \cos A \sin \theta \cos \theta = \sin A \cos^2 \theta + \frac{1}{2} \cos A \sin 2\theta</math>. | ||
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<cmath>620 (\sin 90^\circ + \sin A) = 620 \cdot \frac{6}{5} = \boxed{744}</cmath>. | <cmath>620 (\sin 90^\circ + \sin A) = 620 \cdot \frac{6}{5} = \boxed{744}</cmath>. | ||
− | + | ==Solution 4== | |
Let <math>\alpha</math> be the angle <math>\angle CAS</math> and <math>\beta</math> be the angle <math>\angle BAQ</math>. Then | Let <math>\alpha</math> be the angle <math>\angle CAS</math> and <math>\beta</math> be the angle <math>\angle BAQ</math>. Then | ||
<cmath>\alpha + \beta + \angle A = 90^\circ \Rightarrow \alpha + \beta = 90^\circ - \angle A</cmath> | <cmath>\alpha + \beta + \angle A = 90^\circ \Rightarrow \alpha + \beta = 90^\circ - \angle A</cmath> | ||
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− | + | ==Note on Problem Validity== | |
It has been noted that this answer won't actually work. Let angle <math>QAB = m</math> and angle <math>CAS = n</math> as in Solution 1. Since we know (through that solution) that <math>m = n</math>, we can call them each <math>\theta</math>. The height of the rectangle is <math>AS = 31\cos\theta</math>, and the distance <math>BQ = 40\sin\theta</math>. We know that, if the triangle is to be inscribed in a rectangle, <math>AS \geq BQ</math>. | It has been noted that this answer won't actually work. Let angle <math>QAB = m</math> and angle <math>CAS = n</math> as in Solution 1. Since we know (through that solution) that <math>m = n</math>, we can call them each <math>\theta</math>. The height of the rectangle is <math>AS = 31\cos\theta</math>, and the distance <math>BQ = 40\sin\theta</math>. We know that, if the triangle is to be inscribed in a rectangle, <math>AS \geq BQ</math>. | ||
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label("$S$",S,NW); | label("$S$",S,NW); | ||
</asy> | </asy> | ||
+ | |||
The actual answer is a radical near <math>728</math> (letting the triangle be inside the rectangle). The CAMC, however, has decided to accept only the answer <math>744</math> despite the invalid problem statement. | The actual answer is a radical near <math>728</math> (letting the triangle be inside the rectangle). The CAMC, however, has decided to accept only the answer <math>744</math> despite the invalid problem statement. | ||
− | =See Also= | + | ==See Also== |
{{AIME box|year=2016|n=I|num-b=8|num-a=10}} | {{AIME box|year=2016|n=I|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 15:52, 4 September 2021
Contents
Problem 9 (WARNING: CORRUPTED PROBLEM)
Triangle has and . This triangle is inscribed in rectangle with on and on . Find the maximum possible area of .
Solution 1
Note that if angle is obtuse, it would be impossible for the triangle to inscribed in a rectangle. This can easily be shown by drawing triangle ABC, where is obtuse. Therefore, angle A is acute. Let angle and angle . Then, and . Then the area of rectangle is . By product-to-sum, . . The maximum possible value of is 1, which occurs when . Thus the maximum possible value of is so the maximum possible area of is .
Solution 2
As above, we note that angle must be acute. Therefore, let be the origin, and suppose that is on the positive axis and is on the positive axis. We approach this using complex numbers. Let , and let be a complex number with , and . Then we represent by and by . The coordinates of and depend on the real part of and the imaginary part of . Thus We can expand this, using the fact that , finding Now as , we know that . Also, , so the maximum possible imaginary part of is . This is clearly achievable under our conditions on . Therefore, the maximum possible area of is .
Solution 3 (With Calculus)
Let be the angle . The height of the rectangle then can be expressed as , and the length of the rectangle can be expressed as . The area of the rectangle can then be written as a function of , . For now, we will ignore the and focus on the function .
Taking the derivative, . Setting this equal to , we get . Since we know that , the solution is extraneous. Thus, we get that .
Plugging this value into the original area equation, . Using a product-to-sum formula, we get that: .
Solution 4
Let be the angle and be the angle . Then However, by AM-GM: Therefore, So, . However, the area of the rectangle is just .
Note on Problem Validity
It has been noted that this answer won't actually work. Let angle and angle as in Solution 1. Since we know (through that solution) that , we can call them each . The height of the rectangle is , and the distance . We know that, if the triangle is to be inscribed in a rectangle, .
However, , so the triangle does not actually fit in the rectangle: specifically, B is above R and thus in the line containing segment QR but not on the actual segment or in the rectangle.
The actual answer is a radical near (letting the triangle be inside the rectangle). The CAMC, however, has decided to accept only the answer despite the invalid problem statement.
See Also
2016 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.