Difference between revisions of "2016 AIME I Problems/Problem 9"
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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>. | ||
==Solution== | ==Solution== | ||
+ | |||
+ | ===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>. Since <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>. Since <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>. | ||
-AkashD | -AkashD | ||
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+ | ===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 | ||
+ | <cmath>[AQRS]=\Re(40z)\cdot \Im(31zw)=1240\left(\frac{z+\overline{z}}{2}\right)\left(\frac{zw-\overline{zw}}{2i}\right).</cmath> | ||
+ | We can expand this, using the fact that <math>z\overline{z}=|z|^2</math>, finding | ||
+ | <cmath>[AQRS]=620\left(\frac{z^2w-\overline{z^2w}+w-\overline{w}}{2i}\right)=620(\Im(z^2w)+\Im(w)).</cmath> | ||
+ | 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>. | ||
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=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}} |
Revision as of 04:39, 5 March 2016
Problem
Triangle has and . This triangle is inscribed in rectangle with on and on . Find the maximum possible area of .
Solution
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, . Since . The maximum possible value of is 1, which occurs when . Thus the maximum possible value of is so the maximum possible area of is . -AkashD
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 .
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 |
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