Difference between revisions of "1993 AIME Problems/Problem 13"
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− | Draw a line from point <math>O</math> to <math>Q</math> such that <math>OQ</math> is perpendicular to <math>BD</math>. Further, draw a line passing through points <math>O</math> and <math>P</math>, so <math>OP</math> is parallel to <math>AB</math> and <math>CD</math> and is midway between those two lines. Then <math>PR=\dfrac{AB+CD}{2}=\dfrac{3t+t}{2}=2t</math>. Draw another line passing through point <math>D</math> and parallel to <math>AC</math>, | + | Draw a line from point <math>O</math> to <math>Q</math> such that <math>OQ</math> is perpendicular to <math>BD</math>. Further, draw a line passing through points <math>O</math> and <math>P</math>, so <math>OP</math> is parallel to <math>AB</math> and <math>CD</math> and is midway between those two lines. Then <math>PR=\dfrac{AB+CD}{2}=\dfrac{3t+t}{2}=2t</math>. Draw another line passing through point <math>D</math> and parallel to <math>AC</math>, and call the point of intersection of this line with <math>AB</math> as <math>S</math>. Then <math>SB=AB-AS=3t-t=2t</math>. |
− | We see that <math>m\angle SBD=m\angle ORQ</math> since they are corresponding angles, and thus by angle-angle similarity, <math>\triangle QOR | + | We see that <math>m\angle SBD=m\angle ORQ</math> since they are corresponding angles, and thus by angle-angle similarity, <math>\triangle QOR\sim\triangle SDB</math>. |
Then | Then | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | \dfrac{OQ}{DS}=\dfrac{ | + | \dfrac{OQ}{DS}=\dfrac{RO}{BD}&\implies\dfrac{50}{200}=\dfrac{RO}{\sqrt{200^2+4t^2}}\\ |
− | &\implies | + | &\implies RO=\dfrac{1}{4}\left(\sqrt{200^2+4t^2}\right)\\ |
− | &\implies | + | &\implies RO=\dfrac{1}{2}\left(\sqrt{100^2+t^2}\right) |
\end{align*}</cmath> | \end{align*}</cmath> | ||
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And we obtain | And we obtain | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | PR&= | + | PR-OP&=RO\\ |
− | 2t&= | + | 2t-50&=\dfrac{1}{2}\left(\sqrt{100^2+t^2}\right)\\ |
4t-100&=\sqrt{100^2+t^2}\\ | 4t-100&=\sqrt{100^2+t^2}\\ | ||
(4t-100)^2&=\left(\sqrt{100^2+t^2}\right)^2\\ | (4t-100)^2&=\left(\sqrt{100^2+t^2}\right)^2\\ | ||
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so we have <math>t=\frac{160}{3}</math>, and our answer is thus <math>160+3=\boxed{163}</math>. | so we have <math>t=\frac{160}{3}</math>, and our answer is thus <math>160+3=\boxed{163}</math>. | ||
+ | ===Solution 4=== | ||
+ | |||
+ | We can use areas to find the answer. Since Jenny and Kenny are 200 meters apart, we know that they are side by side, and that the line connecting the two of them is tangent to the circular building. | ||
+ | |||
+ | We know that the radius of the circle is 50, and that <math>\overline{AJ} = x</math>, <math>\overline{BK} = 3x</math>. | ||
+ | |||
+ | Illustration: | ||
+ | |||
+ | <asy> | ||
+ | size(8cm); | ||
+ | pair A = (0,0), B = (200,0), K = (200,160), J = (0,53.3333), O = (100,50); | ||
+ | pair D = tangent(J,O,50,2), F = tangent(A,O,50); | ||
+ | pair[] p = {A,B,K,J,O,D,F}; | ||
+ | for (pair point : p) { | ||
+ | dot(point); | ||
+ | } | ||
+ | label("A",A,SW); | ||
+ | label("B",B,SE); | ||
+ | label("K",K,NE); | ||
+ | label("J",J,NW); | ||
+ | label("O",O,SW); | ||
+ | label("D",D,N); | ||
+ | label("F",F,S); | ||
+ | draw(Circle(O,50)); | ||
+ | draw(A--B--K--J--cycle); | ||
+ | draw(F--O--J); | ||
+ | draw(D--O); | ||
+ | draw(O--K); | ||
+ | label("$50$",(100,26.666),0.5E); | ||
+ | label("$50$",midpoint(D--O),0.5NE); | ||
+ | label("$x$",midpoint(A--J),0.5W); | ||
+ | label("$3x$",midpoint(B--K),0.5E); | ||
+ | label("$200$",midpoint(A--B),4S); | ||
+ | draw(rightanglemark(J,A,F,150)); | ||
+ | draw(rightanglemark(J,D,O,150)); | ||
+ | draw(rightanglemark(A,F,O,150)); | ||
+ | draw(rightanglemark(F,B,K,150)); | ||
+ | </asy> | ||
+ | |||
+ | By areas, <math>[OJK] + [AJOF] + [OFBK] = [ABKJ]</math>. Having right trapezoids, <math>[AFOJ] = \frac{x+50}{2} \cdot 100</math>. The other areas of right trapezoids can be calculated in the same way. We just need to find <math>[OJK]</math> in terms of <math>x</math>. | ||
+ | |||
+ | If we bring <math>\overline{AB}</math> up to where the point J is, we have by the Pythagorean Theorem, <math>\overline{JK} = 2\sqrt{x^2+10000} \Rightarrow [OJK] = \frac{1}{2} \overline{JK} \cdot \overline{OD}</math>. | ||
+ | |||
+ | Now we have everything to solve for <math>x</math>. | ||
+ | |||
+ | <cmath>[OJK] + [AJOF] + [OFBK] = [ABKJ]</cmath> | ||
+ | |||
+ | <cmath>50 \sqrt{x^2 + 10000} + \frac{x+50}{2} \cdot 100 + \frac{3x+50}{2} \cdot 100 = \frac{x+3x}{2} \cdot 200</cmath> | ||
+ | |||
+ | After isolating the radical, dividing by 50, and squaring, we obtain: <math>15x^2 - 800x = 0 \Rightarrow x = \frac{160}{3}</math>. | ||
+ | |||
+ | Since Jenny walks <math>\frac{160}{3}</math> meters at 1 m/s, our answer is <math>160+3 = \fbox{163}</math>. | ||
+ | |||
+ | === Solution 5 === | ||
+ | Basically, draw out a good diagram, and the rest is done. | ||
== See also == | == See also == |
Latest revision as of 01:08, 4 May 2020
Problem
Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and Kenny, they are 200 feet apart. Let be the amount of time, in seconds, before Jenny and Kenny can see each other again. If is written as a fraction in lowest terms, what is the sum of the numerator and denominator?
Contents
Solution
Solution 1
Consider the unit cicle of radius 50. Assume that they start at points and Then at time , they end up at points and The equation of the line connecting these points and the equation of the circle are When they see each other again, the line connecting the two points will be tangent to the circle at the point Since the radius is perpendicular to the tangent we get or Now substitute into and get Now substitute this and into and solve for to get Finally, the sum of the numerator and denominator is
Solution 2
Let and be Kenny's initial and final points respectively and define and similarly for Jenny. Let be the center of the building. Also, let be the intersection of and . Finaly, let and be the points of tangency of circle to and respectively.
From the problem statement, , and . Since , .
Since , . So, .
Since circle is tangent to and , is the angle bisector of .
Thus, .
Therefore, , and the answer is .
Solution 3
Let be the time they walk. Then and .
Draw a line from point to such that is perpendicular to . Further, draw a line passing through points and , so is parallel to and and is midway between those two lines. Then . Draw another line passing through point and parallel to , and call the point of intersection of this line with as . Then .
We see that since they are corresponding angles, and thus by angle-angle similarity, .
Then
And we obtain
so we have , and our answer is thus .
Solution 4
We can use areas to find the answer. Since Jenny and Kenny are 200 meters apart, we know that they are side by side, and that the line connecting the two of them is tangent to the circular building.
We know that the radius of the circle is 50, and that , .
Illustration:
By areas, . Having right trapezoids, . The other areas of right trapezoids can be calculated in the same way. We just need to find in terms of .
If we bring up to where the point J is, we have by the Pythagorean Theorem, .
Now we have everything to solve for .
After isolating the radical, dividing by 50, and squaring, we obtain: .
Since Jenny walks meters at 1 m/s, our answer is .
Solution 5
Basically, draw out a good diagram, and the rest is done.
See also
1993 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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