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Difference between revisions of "1965 AHSME Problems/Problem 36"

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\textbf{(E) }\ \frac{a^2}{a-b} </math>
 
\textbf{(E) }\ \frac{a^2}{a-b} </math>
  
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== Solution ==
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== Solution 1 ==
  
 
<asy>
 
<asy>
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For simplicity, let the first perpendicular from <math>\overleftrightarrow{OA}</math> to <math>\overleftrightarrow{OB}</math> be <math>\overline{AB}</math>, and let the second perpendicular have foot <math>C</math> on <math>\overleftrightarrow{OA}</math>. Further, let the perpendicular from <math>C</math> to <math>\overleftrightarrow{OB}</math> have foot <math>D</math> and length <math>c</math>, as in the diagram. Also, let <math>\measuredangle OAB=\alpha</math>. From the problem, we have <math>AB=a</math> and <math>BC=b</math>. By [[AA similarity]], we have <math>\triangle OCB \sim \triangle OBA</math>, so <math>\measuredangle CBO=\alpha</math> as well. In <math>\triangle ABC</math>, we see that <math>\sin\alpha=\frac{b}{a}</math>, and, in <math>\triangle CDB</math>, <math>\sin\alpha=\frac{c}{b}</math>. Equating these two expressions for <math>\sin\alpha</math>, we get that <math>\frac{b}{a}=\frac{c}{b}</math>, or, because <math>a,b,c>0</math>, <math>b=\sqrt{ac}</math>. Thus, <math>b</math> is the [[geometric mean]] of <math>a</math> and <math>c</math>. Note that if we remove the first perpendicular (i.e. the one with length <math>a</math>), we are left with a smaller version of the original problem, which will have the same equation for the limit (but this time expressed in terms of <math>b</math> and <math>c</math> rather than <math>a</math> and <math>b</math>). Thus, if we let the length of the fourth perpendicular be <math>d</math>, then <math>c</math> will equal the geometric mean of <math>b</math> and <math>d</math>, and so on for the infinitude of perpendiculars. Thus, because the length of a given perpendicular (except the first one) is the geometric mean of the two adjacent perpendiculars, the lengths of the perpendiculars form a [[geometric sequence]]. Because the sequence's first two terms are <math>a</math> and <math>b</math>, it has common ratio <math>\frac{b}{a}</math>. Because <math>b<a</math>, the common ratio has an absolute value less than <math>1</math>, so the sequence's infinite [[geometric sequence#Sum|geometric series]] converges. This infinite sum is given by <math>\frac{a}{1-\frac{b}{a}}=\boxed{\frac{a^2}{a-b}}</math>, which is answer choice <math>\fbox{\textbf{(E)}}</math>.
 
For simplicity, let the first perpendicular from <math>\overleftrightarrow{OA}</math> to <math>\overleftrightarrow{OB}</math> be <math>\overline{AB}</math>, and let the second perpendicular have foot <math>C</math> on <math>\overleftrightarrow{OA}</math>. Further, let the perpendicular from <math>C</math> to <math>\overleftrightarrow{OB}</math> have foot <math>D</math> and length <math>c</math>, as in the diagram. Also, let <math>\measuredangle OAB=\alpha</math>. From the problem, we have <math>AB=a</math> and <math>BC=b</math>. By [[AA similarity]], we have <math>\triangle OCB \sim \triangle OBA</math>, so <math>\measuredangle CBO=\alpha</math> as well. In <math>\triangle ABC</math>, we see that <math>\sin\alpha=\frac{b}{a}</math>, and, in <math>\triangle CDB</math>, <math>\sin\alpha=\frac{c}{b}</math>. Equating these two expressions for <math>\sin\alpha</math>, we get that <math>\frac{b}{a}=\frac{c}{b}</math>, or, because <math>a,b,c>0</math>, <math>b=\sqrt{ac}</math>. Thus, <math>b</math> is the [[geometric mean]] of <math>a</math> and <math>c</math>. Note that if we remove the first perpendicular (i.e. the one with length <math>a</math>), we are left with a smaller version of the original problem, which will have the same equation for the limit (but this time expressed in terms of <math>b</math> and <math>c</math> rather than <math>a</math> and <math>b</math>). Thus, if we let the length of the fourth perpendicular be <math>d</math>, then <math>c</math> will equal the geometric mean of <math>b</math> and <math>d</math>, and so on for the infinitude of perpendiculars. Thus, because the length of a given perpendicular (except the first one) is the geometric mean of the two adjacent perpendiculars, the lengths of the perpendiculars form a [[geometric sequence]]. Because the sequence's first two terms are <math>a</math> and <math>b</math>, it has common ratio <math>\frac{b}{a}</math>. Because <math>b<a</math>, the common ratio has an absolute value less than <math>1</math>, so the sequence's infinite [[geometric sequence#Sum|geometric series]] converges. This infinite sum is given by <math>\frac{a}{1-\frac{b}{a}}=\boxed{\frac{a^2}{a-b}}</math>, which is answer choice <math>\fbox{\textbf{(E)}}</math>.
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== Solution 2 ==
  
 
== See Also ==
 
== See Also ==

Revision as of 20:23, 19 July 2024

Problem

Given distinct straight lines $OA$ and $OB$. From a point in $OA$ a perpendicular is drawn to $OB$; from the foot of this perpendicular a line is drawn perpendicular to $OA$. From the foot of this second perpendicular a line is drawn perpendicular to $OB$; and so on indefinitely. The lengths of the first and second perpendiculars are $a$ and $b$, respectively. Then the sum of the lengths of the perpendiculars approaches a limit as the number of perpendiculars grows beyond all bounds. This limit is:

$\textbf{(A)}\ \frac {b}{a - b} \qquad \textbf{(B) }\ \frac {a}{a - b} \qquad \textbf{(C) }\ \frac {ab}{a - b} \qquad \textbf{(D) }\ \frac{b^2}{a-b}\qquad \textbf{(E) }\ \frac{a^2}{a-b}$

Solution 1

[asy] import geometry; point O=(0,0); point A=(10,5); point B=(10,0); point C; point D; line OA=line(O,A); line OB=line(O,B); // Lines OA and OB draw(OA); draw(OB); // Points O, A, and B dot(O); label("O",O,S); dot(A); label("A",A,NW); dot(B); label("B",B,S); // Segments AB, BC, and CD draw(A--B); pair[] x=intersectionpoints(perpendicular(B,OA),(O--A)); C=x[0]; dot(C); label("C", C, NW); draw(B--C); pair[] y=intersectionpoints(perpendicular(C,OB), (O--B)); D=y[0]; dot(D); label("D", D, S); draw(C--D); // Right Angle Markers markscalefactor=0.075; draw(rightanglemark(O,B,A)); draw(rightanglemark(B,C,O)); draw(rightanglemark(O,D,C)); // Alpha Labels markscalefactor=0.15; draw(anglemark(O,A,B)); draw(anglemark(C,B,O)); label("$\alpha$", (9.6,4.4)); // Length Labels label("$a$", midpoint(A--B), E); label("$b$", midpoint(B--C), E); label("$c$", midpoint(C--D), W); [/asy]

For simplicity, let the first perpendicular from $\overleftrightarrow{OA}$ to $\overleftrightarrow{OB}$ be $\overline{AB}$, and let the second perpendicular have foot $C$ on $\overleftrightarrow{OA}$. Further, let the perpendicular from $C$ to $\overleftrightarrow{OB}$ have foot $D$ and length $c$, as in the diagram. Also, let $\measuredangle OAB=\alpha$. From the problem, we have $AB=a$ and $BC=b$. By AA similarity, we have $\triangle OCB \sim \triangle OBA$, so $\measuredangle CBO=\alpha$ as well. In $\triangle ABC$, we see that $\sin\alpha=\frac{b}{a}$, and, in $\triangle CDB$, $\sin\alpha=\frac{c}{b}$. Equating these two expressions for $\sin\alpha$, we get that $\frac{b}{a}=\frac{c}{b}$, or, because $a,b,c>0$, $b=\sqrt{ac}$. Thus, $b$ is the geometric mean of $a$ and $c$. Note that if we remove the first perpendicular (i.e. the one with length $a$), we are left with a smaller version of the original problem, which will have the same equation for the limit (but this time expressed in terms of $b$ and $c$ rather than $a$ and $b$). Thus, if we let the length of the fourth perpendicular be $d$, then $c$ will equal the geometric mean of $b$ and $d$, and so on for the infinitude of perpendiculars. Thus, because the length of a given perpendicular (except the first one) is the geometric mean of the two adjacent perpendiculars, the lengths of the perpendiculars form a geometric sequence. Because the sequence's first two terms are $a$ and $b$, it has common ratio $\frac{b}{a}$. Because $b<a$, the common ratio has an absolute value less than $1$, so the sequence's infinite geometric series converges. This infinite sum is given by $\frac{a}{1-\frac{b}{a}}=\boxed{\frac{a^2}{a-b}}$, which is answer choice $\fbox{\textbf{(E)}}$.

Solution 2

See Also

1965 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 35 		Followed by
Problem 37
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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