View source for 1965 AHSME Problems/Problem 36

== Problem ==

Given distinct straight lines <math>OA</math> and <math>OB</math>. From a point in <math>OA</math> a perpendicular is drawn to <math>OB</math>; 
from the foot of this perpendicular a line is drawn perpendicular to <math>OA</math>. 
From the foot of this second perpendicular a line is drawn perpendicular to <math>OB</math>; 
and so on indefinitely. The lengths of the first and second perpendiculars are <math>a</math> and <math>b</math>, 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: 

<math>\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} </math>

== 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 <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, intuition) ==
Let <math>a</math> and <math>b</math> 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 <math>\frac{\text{meters}}{\text{meters}}</math>, so they are dimensionless. Unfortunately, our other three options all have units of meters. Now, we have a <math>\frac{1}{3}</math> chance of guessing the answer correctly, but we can go further. Think of what happens as <math>\measuredangle BOA</math> approaches <math>0^{\circ}</math>. Then, <math>a</math> approaches <math>0</math>, 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 <math>\measuredangle BOA</math> approaches <math>90^{\circ}</math>, all of the perpendiculars except the first one (with length <math>a</math>) go to zero. From this intuition, one would think that <math>a</math> 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 <math>\boxed{\textbf{(E) }\frac{a^2}{a-b}}</math>. To reinforce this decision, think about what happens as <math>\measuredangle BOA</math> approaches <math>90^{\circ}</math> again. We would ''expect'' the sum to approach <math>a</math> and <math>b</math> to approach <math>0</math>. Plugging in <math>b=0</math> into the expression in choice (E), we get <math>\frac{a^2}{a}=a</math>, which is what we expected. On the other hand, options (C) and (D) equate to 0, which is clearly false.

== See Also ==
{{AHSME 40p box|year=1965|num-b=35|num-a=37}}
{{MAA Notice}}
[[Category:Intermediate Geometry Problems]]