Difference between revisions of "1972 AHSME Problems/Problem 30"
m (latex fail) |
m (Latex) |
||
Line 46: | Line 46: | ||
Let the rectangle <math>ABCD</math> have crease <math>BE</math> with <math>E</math> on <math>CD</math>, and let <math>F</math> be on <math>AD</math> such that <math>F</math> is a reflection of <math>C</math> over <math>BE</math>. Notice that triangles <math>ABF</math> and <math>DEF</math> are similar, so by setting <math>CE = EF = x</math> with <math>DE = 6-x</math>, giving <math>DF = 2\sqrt{3x-9}</math> we have that <math>AF = \frac{18-3x}{\sqrt{3x-9}}</math>. Noticing that | Let the rectangle <math>ABCD</math> have crease <math>BE</math> with <math>E</math> on <math>CD</math>, and let <math>F</math> be on <math>AD</math> such that <math>F</math> is a reflection of <math>C</math> over <math>BE</math>. Notice that triangles <math>ABF</math> and <math>DEF</math> are similar, so by setting <math>CE = EF = x</math> with <math>DE = 6-x</math>, giving <math>DF = 2\sqrt{3x-9}</math> we have that <math>AF = \frac{18-3x}{\sqrt{3x-9}}</math>. Noticing that | ||
− | Note from xiej: <math>BC=BF=x\cot{\theta}</math>, not <math>x\tan{\theta}</math>, so the solution goes off rails here. | + | <math>\text{\color{red}Note from xiej}</math>: <math>BC=BF=x\cot{\theta}</math>, not <math>x\tan{\theta}</math>, so the solution goes off rails here. |
− | + | ||
+ | <math>BC = BF = x\tan{\theta}</math> gives <math>\frac{(18-3x)^2}{3x-9}+36 = x^2\tan^2{\theta} \Rightarrow \frac{3(x-6)^2}{x-3}+36 = \frac{3x^2}{x-3} = x^2\tan^2{\theta}</math>. <math>x = \frac{3\tan^2{\theta}+3}{\tan^2{\theta}} = \frac{3\sec^2{\theta}}{\tan^2{\theta}} = 3\csc^2{\theta}</math>. Noticing that <math>BE = x\sqrt{\tan^2{\theta}+1} = x\sec{\theta}</math> gives the answer to be <math>3\csc^2{\theta}\sec{\theta}</math> which is not in the answer list? | ||
Revision as of 21:51, 22 August 2017
Problem 30
A rectangular piece of paper 6 inches wide is folded as in the diagram so that one corner touches the opposite side. The length in inches of the crease L in terms of angle is
Solution
Let the rectangle have crease with on , and let be on such that is a reflection of over . Notice that triangles and are similar, so by setting with , giving we have that . Noticing that
: , not , so the solution goes off rails here.
gives . . Noticing that gives the answer to be which is not in the answer list?
If anyone could correct my solution, please help and thanks!
P.S correct answer is .
1972 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 29 |
Followed by Problem 31 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.