Difference between revisions of "1965 AHSME Problems/Problem 36"
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import geometry; | import geometry; | ||
point O=(0,0); | point O=(0,0); | ||
− | point A=(10, | + | point A=(10,5); |
point B=(10,0); | point B=(10,0); | ||
point C; | point C; | ||
+ | point D; | ||
line OA=line(O,A); | line OA=line(O,A); | ||
line OB=line(O,B); | line OB=line(O,B); | ||
Line 37: | Line 38: | ||
label("B",B,S); | label("B",B,S); | ||
− | // Segments AB and | + | // Segments AB, BC, and CD |
draw(A--B); | draw(A--B); | ||
pair[] x=intersectionpoints(perpendicular(B,OA),(O--A)); | pair[] x=intersectionpoints(perpendicular(B,OA),(O--A)); | ||
Line 44: | Line 45: | ||
label("C", C, NW); | label("C", C, NW); | ||
draw(B--C); | 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 | // Right Angle Markers | ||
− | markscalefactor=0. | + | markscalefactor=0.075; |
draw(rightanglemark(O,B,A)); | draw(rightanglemark(O,B,A)); | ||
draw(rightanglemark(B,C,O)); | 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 | // Length Labels | ||
label("$a$", midpoint(A--B), E); | label("$a$", midpoint(A--B), E); | ||
− | label("$b$", midpoint(B--C), | + | label("$b$", midpoint(B--C), E); |
+ | label("$c$", midpoint(C--D), W); | ||
</asy> | </asy> | ||
− | <math>\fbox{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>. |
== See Also == | == See Also == |
Revision as of 20:06, 19 July 2024
Problem
Given distinct straight lines and
. From a point in
a perpendicular is drawn to
;
from the foot of this perpendicular a line is drawn perpendicular to
.
From the foot of this second perpendicular a line is drawn perpendicular to
;
and so on indefinitely. The lengths of the first and second perpendiculars are
and
, 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:
Solution
For simplicity, let the first perpendicular from to
be
, and let the second perpendicular have foot
on
. Further, let the perpendicular from
to
have foot
and length
, as in the diagram. Also, let
. From the problem, we have
and
. By AA similarity, we have
, so
as well. In
, we see that
, and, in
,
. Equating these two expressions for
, we get that
, or, because
,
. Thus,
is the geometric mean of
and
. Note that if we remove the first perpendicular (i.e. the one with length
), 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
and
rather than
and
). Thus, if we let the length of the fourth perpendicular be
, then
will equal the geometric mean of
and
, 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
and
, it has common ratio
. Because
, the common ratio has an absolute value less than
, so the sequence's infinite geometric series converges. This infinite sum is given by
, which is answer choice
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.