Difference between revisions of "2019 AMC 12B Problems/Problem 12"
(→Solution 2) |
m (minor edit) |
||
(15 intermediate revisions by 6 users not shown) | |||
Line 1: | Line 1: | ||
− | ==Problem== | + | == Problem == |
+ | Right triangle <math>ACD</math> with right angle at <math>C</math> is constructed outwards on the hypotenuse <math>\overline{AC}</math> of isosceles right triangle <math>ABC</math> with leg length <math>1</math>, as shown, so that the two triangles have equal perimeters. What is <math>\sin(2\angle BAD)</math>? | ||
− | |||
<asy> | <asy> | ||
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ | /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ | ||
Line 9: | Line 9: | ||
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 */ | ||
− | |||
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.)); | ||
Line 36: | Line 35: | ||
<math>\textbf{(A) } \dfrac{1}{3} \qquad\textbf{(B) } \dfrac{\sqrt{2}}{2} \qquad\textbf{(C) } \dfrac{3}{4} \qquad\textbf{(D) } \dfrac{7}{9} \qquad\textbf{(E) } \dfrac{\sqrt{3}}{2}</math> | <math>\textbf{(A) } \dfrac{1}{3} \qquad\textbf{(B) } \dfrac{\sqrt{2}}{2} \qquad\textbf{(C) } \dfrac{3}{4} \qquad\textbf{(D) } \dfrac{7}{9} \qquad\textbf{(E) } \dfrac{\sqrt{3}}{2}</math> | ||
+ | == Solutions == | ||
+ | === Solution 1 === | ||
+ | Firstly, note by the Pythagorean Theorem in <math>\triangle ABC</math> that <math>AC = \sqrt{2}</math>. Now, the equal perimeter condition means that <math>BC + BA = 2 = CD + DA</math>, since side <math>AC</math> is common to both triangles and thus can be discounted. This relationship, in combination with the Pythagorean Theorem in <math>\triangle ACD</math>, gives <math>AC^2+CD^2=\left(\sqrt{2}\right)^2+\left(2-DA\right)^2=DA^2</math>. Hence <math>2 + 4 - 4DA + DA^2 = DA^2</math>, so <math>DA = \frac{3}{2}</math>, and thus <math>CD = \frac{1}{2}</math>. | ||
− | + | Next, since <math>\angle BAC = 45^{\circ}</math>, <math>\sin{\left(\angle BAC\right)} = \cos{\left(\angle BAC\right)} = \frac{1}{\sqrt{2}}</math>. Using the lengths found above, <math>\sin{\left(\angle CAD\right)} = \frac{\left(\frac{1}{2}\right)}{\left(\frac{3}{2}\right)} = \frac{1}{3}</math>, and <math>\cos{\left(\angle CAD\right)} = \frac{\sqrt{2}}{\left(\frac{3}{2}\right)} = \frac{2 \sqrt{2}}{3}</math>. | |
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
+ | Thus, by the addition formulae for <math>\sin</math> and <math>\cos</math>, we have | ||
+ | <cmath>\begin{split}\sin{\left(\angle BAD\right)}&=\sin{\left(\angle BAC + \angle CAD\right)}\\&=\sin{\left(\angle BAC\right)}\cos{\left(\angle CAD\right)}+\cos{\left(\angle BAC\right)}\sin{\left(\angle CAD\right)}\\&=\frac{1}{\sqrt{2}}\cdot\frac{2 \sqrt{2}}{3} + \frac{1}{\sqrt{2}}\cdot\frac{1}{3} \\ &= \frac{2 \sqrt{2} + 1}{3 \sqrt{2}}\end{split}</cmath> | ||
and | and | ||
+ | <cmath>\begin{split}\cos{\left(\angle BAD\right)}&=\cos{\left(\angle BAC + \angle CAD\right)}\\&=\cos{\left(\angle BAC\right)}\cos{\left(\angle CAD\right)}-\sin{\left(\angle BAC\right)}\sin{\left(\angle CAD\right)}\\&=\frac{1}{\sqrt{2}}\cdot\frac{2 \sqrt{2}}{3} - \frac{1}{\sqrt{2}}\cdot\frac{1}{3} \\ &= \frac{2 \sqrt{2} - 1}{3 \sqrt{2}}\end{split}</cmath> | ||
− | <math> | + | Hence, by the double angle formula for <math>\sin</math>, <math>\sin{\left(2\angle BAD\right)} = 2\sin{\left(\angle BAD\right)}\cos{\left(\angle BAD\right)} = \frac{2(8-1)}{18} = \boxed{\textbf{(D) } \frac{7}{9}}</math>. |
− | <math> | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | <math>\ | + | === Solution 2 (coordinate geometry) === |
+ | 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>. | ||
− | and | + | Thus we have the lengths <math>DE=1+\frac{1}{2\sqrt{2}}</math> (it is just the <math>y</math>-coordinate) and <math>AE = 1-\frac{1}{2\sqrt{2}}</math>. By simple trigonometry in <math>\triangle DAE</math>, we now find <cmath>\sin{\left(\angle BAD\right)} = \frac{\left(1+\frac{1}{2\sqrt{2}}\right)}{\left(\frac{3}{2}\right)} = \frac{\left(2+\frac{1}{\sqrt{2}}\right)}{3} = \frac{2\sqrt{2}+1}{3\sqrt{2}}</cmath> and <cmath>\cos{\left(\angle BAD\right)} = \frac{\left(1-\frac{1}{2\sqrt{2}}\right)}{\left(\frac{3}{2}\right)} = \frac{\left(2-\frac{1}{\sqrt{2}}\right)}{3} = \frac{2\sqrt{2}-1}{3\sqrt{2}}</cmath> just as before. We can then use the double angle formula (as in Solution 1) to deduce <math>\sin{\left(2\angle BAD\right)} = \boxed{\textbf{(D) } \frac{7}{9}}</math>. |
− | <math>\ | + | === Solution 3 (easier finish to Solution 1) === |
+ | Again, use Pythagorean Theorem to find that <math>AD=\frac{3}{2}</math> and <math>CD=\frac{1}{2}</math>. Let <math>\angle DAC=\theta</math>. Note that we want <cmath>\sin{(90+2\theta)}=\cos{2\theta}</cmath> | ||
+ | which is easy to compute: <cmath>\cos{\theta}=\frac{2\sqrt{2}}{3}\implies \cos{2\theta}=2(\frac{8}{9})-1= \boxed{\textbf{(D) } \frac{7}{9}}</cmath> | ||
− | ==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 01:14, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.