1965 AHSME Problems/Problem 36

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

$[asy] import geometry; point O=(0,0); point A=(10,5); point B=(10,0); point C; point D; line OA=line(O,A); line OB=line(O,B); // Lines OA and OB draw(OA); draw(OB); // Points O, A, and B dot(O); label("O",O,S); dot(A); label("A",A,NW); dot(B); label("B",B,S); // Segments AB, BC, and CD draw(A--B); pair[] x=intersectionpoints(perpendicular(B,OA),(O--A)); C=x[0]; dot(C); label("C", C, NW); draw(B--C); pair[] y=intersectionpoints(perpendicular(C,OB), (O--B)); D=y[0]; dot(D); label("D", D, S); draw(C--D); // Right Angle Markers markscalefactor=0.075; draw(rightanglemark(O,B,A)); draw(rightanglemark(B,C,O)); draw(rightanglemark(O,D,C)); // Alpha Labels markscalefactor=0.15; draw(anglemark(O,A,B)); draw(anglemark(C,B,O)); label("$\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 $\fbox{\textbf{(E)}}$ .

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

Page

Toolbox

Search

1965 AHSME Problems/Problem 36

Problem

Solution

See Also