Difference between revisions of "2019 AMC 12B Problems/Problem 12"
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pen dotstyle = black; /* point style */ | pen dotstyle = black; /* point style */ | ||
real xmin = -4.001920114613276, xmax = 4.014313525192017, ymin = -2.552570341575814, ymax = 5.6249093771911145; /* image dimensions */ | real xmin = -4.001920114613276, xmax = 4.014313525192017, ymin = -2.552570341575814, ymax = 5.6249093771911145; /* image dimensions */ | ||
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draw((-1.6742337260757447,-1.)--(-1.6742337260757445,-0.6742337260757447)--(-2.,-0.6742337260757447)--(-2.,-1.)--cycle, linewidth(2.)); | draw((-1.6742337260757447,-1.)--(-1.6742337260757445,-0.6742337260757447)--(-2.,-0.6742337260757447)--(-2.,-1.)--cycle, linewidth(2.)); | ||
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=== Solution 2 (coordinate geometry) === | === Solution 2 (coordinate geometry) === | ||
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We use the Pythagorean Theorem, as in Solution 1, to find <math>AD=\frac{3}{2}</math> and <math>CD=\frac{1}{2}</math>. Now notice that the angle between <math>CD</math> and the vertical (i.e. the <math>y</math>-axis) is <math>45^{\circ}</math> – to see this, drop a perpendicular from <math>D</math> to <math>BA</math> which meets <math>BA</math> at <math>E</math>, and use the fact that the angle sum of quadrilateral <math>CBED</math> must be <math>360^{\circ}</math>. Anyway, this implies that the line <math>CD</math> has slope <math>1</math>, so since <math>C</math> is the point <math>(0,1)</math> and the length of <math>CD</math> is <math>\frac{1}{2}</math>, <math>D</math> has coordinates <math>\left(0+\frac{\left(\frac{1}{2}\right)}{\sqrt{2}}, 1+\frac{\left(\frac{1}{2}\right)}{\sqrt{2}}\right) = \left(\frac{1}{2\sqrt{2}}, 1+\frac{1}{2\sqrt{2}}\right)</math>. | We use the Pythagorean Theorem, as in Solution 1, to find <math>AD=\frac{3}{2}</math> and <math>CD=\frac{1}{2}</math>. Now notice that the angle between <math>CD</math> and the vertical (i.e. the <math>y</math>-axis) is <math>45^{\circ}</math> – to see this, drop a perpendicular from <math>D</math> to <math>BA</math> which meets <math>BA</math> at <math>E</math>, and use the fact that the angle sum of quadrilateral <math>CBED</math> must be <math>360^{\circ}</math>. Anyway, this implies that the line <math>CD</math> has slope <math>1</math>, so since <math>C</math> is the point <math>(0,1)</math> and the length of <math>CD</math> is <math>\frac{1}{2}</math>, <math>D</math> has coordinates <math>\left(0+\frac{\left(\frac{1}{2}\right)}{\sqrt{2}}, 1+\frac{\left(\frac{1}{2}\right)}{\sqrt{2}}\right) = \left(\frac{1}{2\sqrt{2}}, 1+\frac{1}{2\sqrt{2}}\right)</math>. | ||
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== See Also == | == See Also == | ||
{{AMC12 box|year=2019|ab=B|num-b=11|num-a=13}} | {{AMC12 box|year=2019|ab=B|num-b=11|num-a=13}} | ||
+ | {{MAA Notice}} |
Latest revision as of 00:14, 19 October 2020
Contents
Problem
Right triangle with right angle at is constructed outwards on the hypotenuse of isosceles right triangle with leg length , as shown, so that the two triangles have equal perimeters. What is ?
Solutions
Solution 1
Firstly, note by the Pythagorean Theorem in that . Now, the equal perimeter condition means that , since side is common to both triangles and thus can be discounted. This relationship, in combination with the Pythagorean Theorem in , gives . Hence , so , and thus .
Next, since , . Using the lengths found above, , and .
Thus, by the addition formulae for and , we have and
Hence, by the double angle formula for , .
Solution 2 (coordinate geometry)
We use the Pythagorean Theorem, as in Solution 1, to find and . Now notice that the angle between and the vertical (i.e. the -axis) is – to see this, drop a perpendicular from to which meets at , and use the fact that the angle sum of quadrilateral must be . Anyway, this implies that the line has slope , so since is the point and the length of is , has coordinates .
Thus we have the lengths (it is just the -coordinate) and . By simple trigonometry in , we now find and just as before. We can then use the double angle formula (as in Solution 1) to deduce .
Solution 3 (easier finish to Solution 1)
Again, use Pythagorean Theorem to find that and . Let . Note that we want which is easy to compute:
See Also
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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 |
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