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| | + | {{duplicate|[[2024 AMC 12A Problems/Problem 10|2024 AMC 12A #10]] and [[2024 AMC 10A Problems/Problem 13|2024 AMC 10A #13]]}} |
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| | ==Problem== | | ==Problem== |
| − | Let <math>\alpha</math> be the radian measure of the smallest angle in a <math>3{-}4{-}5</math> right triangle. Let <math>\beta</math> be the radian measure of the smallest angle in a <math>7{-}24{-}25</math> right triangle. In terms of <math>\alpha</math>, what is <math>\beta</math>?
| + | Aubrey raced his younger brother Blair. Aubrey runs at a faster constant speed than Blair, so Blair started the race <math>40</math> feet ahead of Aubrey. Aubrey caught up to Blair after <math>8</math> seconds, finishing the race <math>90</math> feet ahead of Blair and <math>5</math> seconds earlier than Blair. How far did Aubrey run, in feet? |
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| − | <math>\textbf{(A) }\frac{\alpha}{3}\qquad \textbf{(B) }\alpha - \frac{\pi}{8}\qquad \textbf{(C) }\frac{\pi}{2} - 2\alpha \qquad \textbf{(D) }\frac{\alpha}{2}\qquad \textbf{(E) }\pi - 4\alpha\qquad</math>
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| − | | |
| − | ==Solution 1==
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| − | From the question,
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| − | <cmath>\tan\alpha=\frac{3}{4}, \space \tan\beta=\frac{7}{24}</cmath>
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| − | <cmath>\tan(\alpha+\beta)= \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}</cmath>
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| − | <cmath>\tan(\alpha+\beta)= \frac{\frac{3}{4}+\frac{7}{24}}{1-\frac{3}{4} \cdot \frac{7}{24}}</cmath>
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| − | <cmath>\tan(\alpha+\beta)=\frac{4}{3}</cmath>
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| − | <cmath>\alpha+\beta=\tan^{-1}\left(\frac{4}{3}\right)</cmath>
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| − | <cmath>\alpha+\beta=\frac{\pi}{2}-\tan^{-1}\left(\frac{3}{4}\right)</cmath>
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| − | <cmath>\alpha+\beta=\frac{\pi}{2}-\alpha</cmath>
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| − | <cmath>\beta=\boxed{\textbf{(C) }\frac{\pi}{2}-2\alpha}</cmath>
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| − | | |
| − | ~lptoggled
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| − | | |
| − | ==Solution 2: Scaling and combining triangles==
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| − | | |
| − | We can scale the <math>3</math>-<math>4</math>-<math>5</math> triangle up by a factor of <math>6</math> to make its side lengths <math>18,24,</math> and <math>30,</math> then glue its side of length <math>24</math> to the corresponding side in the <math>7</math>-<math>24</math>-<math>25</math> triangle:
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| − | | |
| − | <asy>
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| − | pair A = (0,0);
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| − | pair B = (18,0);
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| − | pair C = (25,0);
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| − | pair D = (18,24);
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| − | | |
| − | draw(A--C--D--cycle);
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| − | draw(B--D);
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| − | | |
| − | draw(rightanglemark(C,B,D,50));
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| − | | |
| − | label("A", A, SW);
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| − | label("B", B, S);
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| − | label("C", C, SE);
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| − | label("D", D, N);
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| − | | |
| − | label("18", A--B, S);
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| − | label("7", B--C, S);
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| − | label("25", C--D, NE);
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| − | label("30", D--A, NW);
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| − | label("24", B--D, W);
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| − | | |
| − | label("$\alpha$", D, 6*dir(250));
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| − | label("$\beta$", D, 9*dir(278));
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| − | label("$\frac{\pi}{2} - \alpha$", A, 4*dir(25));
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| − | </asy>
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| − | Angles <math>\angle DAB</math> and <math>\angle BDA</math> are complementary in <math>\triangle ABD,</math> so <math>\angle DAB = \frac{\pi}{2} - \alpha.</math> We also have <math>AC = 18 + 7 = 25 = CD,</math> so <math>\triangle ACD</math> is isosceles. That means that its base angles <math>\angle CDA</math> and <math>\angle CAD</math> are congruent, so <math>\alpha + \beta = \frac{\pi}{2} - \alpha,</math> and hence <math>\beta = \boxed{\textbf{(C) }\frac{\pi}{2}-2\alpha}.</math>
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| − | | |
| − | ~MartianTom
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| − | | |
| − | ==Solution 3: Trial and Error ==
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| − | Another approach to solving this problem is trial and error, comparing the sine of the answer choices with <math>\sin\beta = \frac{7}{25}</math>. Starting with the easiest sine to compute from the answer choices (option choice D). We get:
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| − | <cmath>\sin{\left(\frac{\alpha}{2}\right)} = \sqrt{\frac{1 - \cos{\alpha}}{2}}</cmath>
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| − | <cmath>= \sqrt{\frac{1 - \frac{4}{5}}{2}}</cmath>
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| − | <cmath>= \sqrt{\frac{1}{10}}</cmath>
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| − | <cmath>\neq \frac{7}{25}</cmath>
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| − | | |
| − | The next easiest sine to compute is option choice C.
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| − | <cmath>\sin{\left(\frac{\pi}{2} - 2\alpha\right)} = \sin{\left(\frac{\pi}{2}\right)}\cos{\left(2\alpha\right)}</cmath>
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| − | <cmath>=\cos{2\alpha}</cmath>
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| − | <cmath>=\cos^2{\alpha} - \sin^2{\alpha}</cmath>
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| − | <cmath>=\frac{16}{25} - \frac{9}{25}</cmath>
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| − | <cmath>=\frac{7}{25}</cmath>
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| − | Since <math>\sin\left(\frac{\pi}{2} - 2\alpha\right)</math> is equal to <math>\sin\beta</math>, option choice C is the correct answer. ~amshah
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| − | | |
| − | ==Solution 4: ==
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| − | <math>sin(B) = \frac{24}{25} = 2 \cdot \frac{12}{25} = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = 2 \cdot sin(A) \cdot cos(A) = sin(2A) = cos(90^{\circ} - 2A)</math> | |
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| − | Therefore <cmath>\beta=\boxed{(C) \frac{\pi}{2} -2\alpha}</cmath>
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| − | | |
| − | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
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| − | | |
| − | ==Solution 5: Ptolemy (no trig)==
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| − | Let AB have length 15, BC have length 20, AC length 25, AD length 7 and CD length 24. Let x be the measure of segment BD. Thus the measure of angle ACB is <math>\alpha</math> and the measure of angle ACD is <math>\beta</math>. ABCD is a cyclic quadrilateral because angle ABC and angle ADC are right angles. Using Ptolemy's theorem on this quadrilateral yields 25x = 15*24 + 7*20 = 500, or x = 20. This means triangle CBD is isoceles. The perpendicular bisector of CD passes through the center (O) of the circle on which ABCD lies and also passes through B. Let the intersection of the perpendicular bisector of CD and CD be point P. The measure of angle OBC is the same as the measure of the angle OCB which is <math>\alpha</math>, so the measure of angle BOC is <math>{\pi} - {2}{\alpha}</math>, so the measure of angle COP is <math>{2}{\alpha}</math>. Triangle COP is a right triangle with angle OCP being the same as angle ACD (<math>\beta</math>), angle COP being <math>{2}{\alpha}</math>, and angle CPO being <math>\frac{\pi}{2}</math>.
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| − | So:
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| − | <cmath>\beta + {2}\alpha + \frac{\pi}{2} = \pi</cmath>
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| − | <cmath>\beta=\boxed{(C) \frac{\pi}{2} -2\alpha}</cmath>~Ilaggo2432
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| − | | |
| − | ==Solution 6(rough value)==
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| − | given in the question,sin(a)is 3/5 and sin (b) is 7/25,estimate the angle by using the arcsin function. arcsin 3/5 is around 0.643 rad and arcsin 7/25 is around 0.283. Note that pi/2 is closest to 1.57 rad, try one by one and option c suggests that 1.57=2(0.643)+0.283. My first amc 12 edit
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| − | ==Solution 7 (Angle Addition)==
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| − | <math>\sin(\alpha)\cos(\beta) + \sin(\beta)\cos(\alpha) = \sin(\alpha+\beta) = \frac{4}{5}</math>. Noticing <math>\frac{4}{5} = \cos(\alpha) = \sin\left(\frac{\pi}{2}-\alpha\right)</math> gives us <math>\alpha + \beta = \frac{\pi}{2} - \alpha</math> so <math>\boxed{\textbf{(C) }\dfrac{\pi}{2} - 2\alpha}</math>
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| − | ~KEVIN_LIU
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| − | | |
| − | ==Solution 8 (Geometry Solution Without Words)==
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| − | <cmath>\angle A+ \angle B+ \angle A = \frac{\pi}{2}</cmath>
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| − | Therefore <cmath>\beta=\boxed{(C) \frac{\pi}{2} -2\alpha}</cmath>
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| − | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
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| − | [[Image:2024_amc12A_p10.PNG|thumb|center|400px|]]
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| − | | |
| − | ==Solution 9 (Complex Number)==
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| − | | |
| − | Define <math>Z_{A} = 4 + 3i = 5e^{iA} </math>
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| − | and <math>Z_{B} = 24 + 7i = 25e^{iB} </math>
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| − | Looking at the answer choices, we start by testing out an easier choice, such as C.
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| − | <math> (4+3i)^2</math>
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| − | <math>= 16 + 9 \cdot i^2 + 2\cdot12 \cdot i</math>
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| − | <math>= 16 - 9 + 24i </math>
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| − | <math>= 7 + 24i </math>
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| − | Finally:
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| − | <math> 25^2 \cdot e^{i(2A+B)} </math>
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| − | <math>= 25e^{i2A} \cdot 25e^{iB} </math>
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| − | <math>={(5e^{iA})}^2 \cdot 25e^{iB}</math>
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| − | <math>= (7+24i)(24+7i)</math>
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| − | <math>=25^2i</math>
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| − | <math>=25^2 e^{i\frac{\pi}{2}}</math>
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| − | This proves that <math>\angle 2A+ \angle B = \frac{\pi}{2}</math>. (Given that <math>\angle A < \frac{\pi}{2}</math> and <math>\angle B < \frac{\pi}{2}</math>)
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| − | Therefore <cmath>\beta=\boxed{(C) \frac{\pi}{2} -2\alpha}</cmath>
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| − | | |
| − | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
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| − | | |
| − | ==Solution 10 (Double Angle + Co-function Identity)==
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| − | <cmath>\cos(\alpha) = \frac{4}{5} \quad \text{and} \quad \sin(\alpha) = \frac{3}{5}</cmath>
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| − | | |
| − | Use a double angle identity to calculate <math>\cos(2\alpha)</math>:
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| − | <cmath>\cos(2\alpha) = \cos^2(\alpha) - \sin^2(\alpha) = \left(\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}</cmath>
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| − | We know that:
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| − | <cmath>\sin(\beta) = \frac{7}{25}</cmath>
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| − | | |
| − | Equate both to each other:
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| − | <cmath>\cos(2\alpha) = \sin(\beta)</cmath>
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| − | | |
| − | Apply the co-function identity, <math>\cos(2\alpha) = \sin\left(\frac{\pi}{2} - 2\alpha\right)</math>:
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| − | <cmath>\sin\left(\frac{\pi}{2} - 2\alpha\right) = \sin(\beta)</cmath>
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| − | | |
| − | Hence,
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| − | <cmath>\beta=\boxed{(\mathbf{C}) \frac{\pi}{2} - 2\alpha}</cmath>
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| − | | |
| − | ~sourodeepdeb
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| − | | |
| − | ==Solution 11 (don't do this)==
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| − | <cmath>\frac{3}{5} \approx \alpha-\frac{\alpha^{3}}{6}</cmath>
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| − | <cmath>-5\alpha^3+30\alpha-18 \approx 0</cmath>
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| − | <cmath>\alpha \approx 0.64464</cmath>
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| − | <cmath>\frac{7}{25} \approx \beta-\frac{\beta^{3}}{6}</cmath>
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| − | <cmath>-25\beta^3+150\beta-42 \approx 0</cmath>
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| − | <cmath>\beta \approx 0.28381</cmath>
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| − | <cmath>\frac{\pi}{2} - 2 \cdot 0.64464 \approx 0.28151632679 \approx \beta</cmath>
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| − | <cmath>\beta=\boxed{(\mathbf{C}) \frac{\pi}{2} - 2\alpha}</cmath>
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| − | | |
| − | ==Solution 12==
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| − | <cmath>\frac{4}{5} \approx 1-\frac{\alpha^2}{2}</cmath>
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| − | <cmath>5\alpha^2 \approx 2</cmath>
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| − | <cmath>\alpha \approx \sqrt{\frac{2}{5}} \approx 0.63246</cmath>
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| − | <cmath>\frac{24}{25} \approx 1-\frac{\beta^2}{2}</cmath>
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| − | <cmath>25\beta^2 \approx 2</cmath>
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| − | <cmath>\beta \approx \sqrt{\frac{2}{25}} \approx 0.28284</cmath>
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| − | <cmath>\frac{\pi}{2} - 2 \cdot 0.63246 \approx 0.30587632679 \approx \beta</cmath>
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| − | <cmath>\beta=\boxed{(\mathbf{C}) \frac{\pi}{2} - 2\alpha}</cmath>
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| − | == Video Solution (🚀 2 min solve 🚀) ==
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| − | https://youtu.be/7UPO4bF5VoI
| + | <math>\textbf{(A)}~454\qquad\textbf{(B)}~494\qquad\textbf{(C)}~518\qquad\textbf{(D)}~558\qquad\textbf{(E)}~598</math> |
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| − | <i>~Education, the Study of Everything</i> | + | ==Solution== |
| | + | Assume the race starts at time <math>t = 0</math>, when Aubrey trails his younger brother Blair by <math>40</math> feet. By time <math>t = 8</math> seconds, Aubrey has "made up" those <math>40</math> feet, so his running speed is exactly <math>\tfrac{40}{8} = 5</math> feet per second faster than Blair. Therefore, if Aubrey finished the race <math>90</math> feet ahead of Blair, he must have done so <math>\tfrac{90}{5} = 18</math> seconds after he caught up. This means Aubrey finished the race at time <math>t = 8 + 18 = 26</math> seconds. Suppose Aubrey ran <math>d</math> feet. Then Blair finished the race at time <math>t = 31</math> seconds and ran <math>d - 40</math> feet. Aubrey was running at a speed <math>5</math> feet per second faster throughout, and therefore <cmath>\frac{d}{26} =\frac{d - 40}{31} + 5 =\frac{d + 115}{31}</cmath> hence <math>31d = 26(d + 115)</math>, so <math>5d = 26 \cdot 115</math> and <math>d = 26 \cdot 23 = \boxed{\textbf{(E)}~598}</math>. |
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| | ==See also== | | ==See also== |
| | {{AMC12 box|year=2024|ab=A|num-b=9|num-a=11}} | | {{AMC12 box|year=2024|ab=A|num-b=9|num-a=11}} |
| | + | {{AMC10 box|year=2024|ab=A|num-b=12|num-a=14}} |
| | {{MAA Notice}} | | {{MAA Notice}} |
feet ahead of Aubrey. Aubrey caught up to Blair after 
seconds, finishing the race 
feet ahead of Blair and 
seconds earlier than Blair. How far did Aubrey run, in feet?
, when Aubrey trails his younger brother Blair by 
seconds, Aubrey has "made up" those 
feet per second faster than Blair. Therefore, if Aubrey finished the race 
seconds after he caught up. This means Aubrey finished the race at time 
seconds. Suppose Aubrey ran 
feet. Then Blair finished the race at time 
seconds and ran 
feet. Aubrey was running at a speed ![\[\frac{d}{26} =\frac{d - 40}{31} + 5 =\frac{d + 115}{31}\]](http://latex.artofproblemsolving.com/5/9/7/597ef0f06747aa3b6407aba3845a4bd604146cee.png)
hence 
, so 
and 
.
