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

Revision as of 20:01, 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 is positive and 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 (Answer choices, non-rigorous intuition)

Let $a$ and $b$ be measured with some units of length (say, meters). The limit of the sum of the lengths of the perpendiculars would, then, be measured in meters as well. Performing dimensional analysis on each of the answer choices, we can eliminate options (A) and (B), because they have units $\frac{\text{meters}}{\text{meters}}$ , so they are dimensionless. Unfortunately, our other three options all have units of meters. Now, we have a $\frac{1}{3}$ chance of guessing the answer correctly, but we can go further. Think of what happens as $\measuredangle BOA$ approaches $0^{\circ}$ . Then, $a$ approaches $0$ , and the perpendiculars have nearly no space to "bounce between" the two lines, and so they likely have zero total length. Likewise, if we think of what happens as $\measuredangle BOA$ approaches $90^{\circ}$ , all of the perpendiculars except the first one (with length $a$ ) go to zero. From this intuition, one would think that $a$ would have a larger impact upon the total length of the perpendiculars. Out of the three answer choices left, this conjecture is only consistent with choice $\boxed{\textbf{(E) }\frac{a^2}{a-b}}$ . To reinforce this decision, think about what happens as $\measuredangle BOA$ approaches $90^{\circ}$ again. We would expect the sum to approach $a$ and $b$ to approach $0$ . Plugging in $b=0$ into the expression in choice (E), we get $\frac{a^2}{a}=a$ , which is what we expected. On the other hand, options (C) and (D) equate to 0, which is clearly false.

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 </asy>
-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 is positive and 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>.
 == Solution 2 (Answer choices, non-rigorous intuition) ==

1965 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 35	Followed by Problem 37
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Difference between revisions of "1965 AHSME Problems/Problem 36"

Revision as of 20:01, 19 July 2024

Contents

Problem

Solution 1

Solution 2 (Answer choices, non-rigorous intuition)

See Also