Difference between revisions of "2021 AIME I Problems/Problem 11"
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Let <math>ABCD</math> be a cyclic quadrilateral with <math>AB=4,BC=5,CD=6,</math> and <math>DA=7</math>. Let <math>A_1</math> and <math>C_1</math> be the feet of the perpendiculars from <math>A</math> and <math>C</math>, respectively, to line <math>BD,</math> and let <math>B_1</math> and <math>D_1</math> be the feet of the perpendiculars from <math>B</math> and <math>D,</math> respectively, to line <math>AC</math>. The perimeter of <math>A_1B_1C_1D_1</math> is <math>\frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | Let <math>ABCD</math> be a cyclic quadrilateral with <math>AB=4,BC=5,CD=6,</math> and <math>DA=7</math>. Let <math>A_1</math> and <math>C_1</math> be the feet of the perpendiculars from <math>A</math> and <math>C</math>, respectively, to line <math>BD,</math> and let <math>B_1</math> and <math>D_1</math> be the feet of the perpendiculars from <math>B</math> and <math>D,</math> respectively, to line <math>AC</math>. The perimeter of <math>A_1B_1C_1D_1</math> is <math>\frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
− | == | + | ==Diagram== |
− | [[File:Leonard_my_dude's_image.png]] | + | [[File:Leonard_my_dude's_image.png|center]] |
+ | ==Solution 1== | ||
Let <math>O</math> be the intersection of <math>AC</math> and <math>BD</math>. Let <math>\theta = \angle AOB</math>. | Let <math>O</math> be the intersection of <math>AC</math> and <math>BD</math>. Let <math>\theta = \angle AOB</math>. | ||
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From Ptolemy's theorem, we have that <math>(BD)(AC) = 4\times6+5\times7 = 59</math>. From Brahmagupta's Formula, <math>[ABCD] = \sqrt{(11-4)(11-5)(11-6)(11-7)} = 2\sqrt{210}</math>. But the area is also <math>\frac{1}{2}(BD)(AC)\sin\theta = \frac{59}{2}\sin\theta</math>, so <math>\sin \theta = \frac{4\sqrt{210}}{59} \implies \cos \theta = \frac{11}{59}</math>. Then the desired fraction is <math>(4+5+6+7)\cos\theta = \frac{242}{59}</math> for an answer of <math>\boxed{301}</math>. | From Ptolemy's theorem, we have that <math>(BD)(AC) = 4\times6+5\times7 = 59</math>. From Brahmagupta's Formula, <math>[ABCD] = \sqrt{(11-4)(11-5)(11-6)(11-7)} = 2\sqrt{210}</math>. But the area is also <math>\frac{1}{2}(BD)(AC)\sin\theta = \frac{59}{2}\sin\theta</math>, so <math>\sin \theta = \frac{4\sqrt{210}}{59} \implies \cos \theta = \frac{11}{59}</math>. Then the desired fraction is <math>(4+5+6+7)\cos\theta = \frac{242}{59}</math> for an answer of <math>\boxed{301}</math>. | ||
− | ==Finding | + | ==Solution 2 (Finding cos{x})== |
− | The angle <math>\theta</math> between diagonals satisfies <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{(s-b)(s-d}{(s-a)(s-c)}}</cmath> (see https://en.wikipedia.org/wiki/Cyclic_quadrilateral#Angle_formulas). | + | The angle <math>\theta</math> between diagonals satisfies <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{(s-b)(s-d)}{(s-a)(s-c)}}</cmath> (see https://en.wikipedia.org/wiki/Cyclic_quadrilateral#Angle_formulas). |
Thus, <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-4)(11-6)}{(11-5)(11-7)}}</cmath> or <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-5)(11-7)}{(11-4)(11-6)}}</cmath> | Thus, <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-4)(11-6)}{(11-5)(11-7)}}</cmath> or <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-5)(11-7)}{(11-4)(11-6)}}</cmath> | ||
That is, <math>\tan^2{\frac{\theta}{2}}=\frac{1-\cos^2{\frac{\theta}{2}}}{\cos^2{\frac{\theta}{2}}}=\frac{24}{35}</math> or <math>\frac{35}{24}</math> | That is, <math>\tan^2{\frac{\theta}{2}}=\frac{1-\cos^2{\frac{\theta}{2}}}{\cos^2{\frac{\theta}{2}}}=\frac{24}{35}</math> or <math>\frac{35}{24}</math> |
Revision as of 00:29, 23 April 2021
Contents
Problem
Let be a cyclic quadrilateral with and . Let and be the feet of the perpendiculars from and , respectively, to line and let and be the feet of the perpendiculars from and respectively, to line . The perimeter of is , where and are relatively prime positive integers. Find .
Diagram
Solution 1
Let be the intersection of and . Let .
Firstly, since , we deduce that is cyclic. This implies that , with a ratio of . This means that . Similarly, . Hence It therefore only remains to find .
From Ptolemy's theorem, we have that . From Brahmagupta's Formula, . But the area is also , so . Then the desired fraction is for an answer of .
Solution 2 (Finding cos{x})
The angle between diagonals satisfies (see https://en.wikipedia.org/wiki/Cyclic_quadrilateral#Angle_formulas). Thus, or That is, or Thus, or In this context, . Thus, ~y.grace.yu
Solution 3 (Pythagorean Theorem)
We assume that the two quadrilateral mentioned in the problem are similar (due to both of them being cyclic). Note that by Ptolemy’s, one of the diagonals has length WLOG we focus on diagonal To find the diagonal of the inner quadrilateral, we drop the altitude from and and calculate the length of Let be (Thus By Pythagorean theorem, we have Now let be (thus making ). Similarly, we have We see that , the scaled down diagonal is just which is times our original diagonal implying a scale factor of Thus, due to perimeters scaling linearly, the perimeter of the new quadrilateral is simply making our answer -fidgetboss_4000
See also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.