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 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 35
Followed by
Problem 37
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