Difference between revisions of "1983 AHSME Problems/Problem 23"
Sevenoptimus (talk | contribs) (Added a solution) |
Sevenoptimus (talk | contribs) m (Improved clarity/formatting and added box at the bottom) |
||
(One intermediate revision by the same user not shown) | |||
Line 35: | Line 35: | ||
[[File:pdfresizer.com-pdf-convert-q23.png]] | [[File:pdfresizer.com-pdf-convert-q23.png]] | ||
− | Consider three consecutive circles, as shown in the diagram above; observe that their centres <math>P</math>, <math>Q</math>, and <math>R</math> are collinear by symmetry. Let <math>A</math>, <math>B</math>, and <math>C</math> be the points of tangency, and let <math>PS</math> and <math>QT</math> be segments parallel to the upper tangent (i.e. <math>L_1</math>), as also shown. Since <math>PQ</math> is parallel to <math>QR</math> (the three points are collinear), <math>PS</math> is parallel to <math>QT</math> (as both are parallel to <math>L_1</math>), and <math>SQ</math> is parallel to <math>TR</math> (as both are perpendicular to <math>L_1</math>), we have <math>\triangle PQS \sim \triangle QRT</math>. | + | Consider three consecutive circles, as shown in the diagram above; observe that their centres <math>P</math>, <math>Q</math>, and <math>R</math> are collinear by symmetry. Let <math>A</math>, <math>B</math>, and <math>C</math> be the points of tangency, and let <math>PS</math> and <math>QT</math> be segments parallel to the upper tangent (i.e. <math>L_1</math>), as also shown. Since <math>PQ</math> is parallel to <math>QR</math> (the three points are collinear), <math>PS</math> is parallel to <math>QT</math> (as both are parallel to <math>L_1</math>), and <math>SQ</math> is parallel to <math>TR</math> (as both are perpendicular to <math>L_1</math>, due to the tangent being perpendicular to the radius), we have <math>\triangle PQS \sim \triangle QRT</math>. |
− | Now, if we let <math>x, y</math>, and <math>z</math> be the radii of the three circles (from smallest to largest), then <math>QS = y - x</math> and <math>RT = z - y</math>. Thus, from the similarity that we just proved, <math>\frac{QS}{PQ} = \frac{RT}{QR} \Rightarrow \frac{y-x}{x+y} = \frac{z-y}{y+z}</math> (where e.g. <math>PQ = x + y</math> because of collinearity). This | + | Now, if we let <math>x, y</math>, and <math>z</math> be the radii of the three circles (from smallest to largest), then <math>QS = y - x</math> and <math>RT = z - y</math>. Thus, from the similarity that we just proved, <math>\frac{QS}{PQ} = \frac{RT}{QR} \Rightarrow \frac{y-x}{x+y} = \frac{z-y}{y+z}</math> (where e.g. <math>PQ = x + y</math> because of collinearity). This equation reduces to <math>y^2 = zx</math>, i.e. <math>\frac{y}{x} = \frac{z}{y}</math>, so the ratio of consecutive radii is constant, forming a geometric sequence. In this case, as the first radius is <math>8</math> and, four radii later, the radius is <math>18</math>, this constant ratio is <math>\sqrt[\leftroot{-2}\uproot{2}{4}]{\frac{18}{8}}</math>. Therefore the middle radius is <math>8 \cdot {\left(\sqrt[\leftroot{-2}\uproot{2}{4}]{\frac{18}{8}}\right)}^{2} = 8 \sqrt{\frac{18}{8}} = \sqrt{18 \cdot 8} = \sqrt{144} = 12</math>, which is choice <math>\boxed{\textbf{(A)}}</math>. |
+ | |||
+ | ==See Also== | ||
+ | {{AHSME box|year=1983|num-b=22|num-a=24}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 01:01, 20 February 2019
Problem
In the adjoining figure the five circles are tangent to one another consecutively and to the lines
and
.
If the radius of the largest circle is
and that of the smallest one is
, then the radius of the middle circle is
Solution
Consider three consecutive circles, as shown in the diagram above; observe that their centres ,
, and
are collinear by symmetry. Let
,
, and
be the points of tangency, and let
and
be segments parallel to the upper tangent (i.e.
), as also shown. Since
is parallel to
(the three points are collinear),
is parallel to
(as both are parallel to
), and
is parallel to
(as both are perpendicular to
, due to the tangent being perpendicular to the radius), we have
.
Now, if we let , and
be the radii of the three circles (from smallest to largest), then
and
. Thus, from the similarity that we just proved,
(where e.g.
because of collinearity). This equation reduces to
, i.e.
, so the ratio of consecutive radii is constant, forming a geometric sequence. In this case, as the first radius is
and, four radii later, the radius is
, this constant ratio is
. Therefore the middle radius is
, which is choice
.
See Also
1983 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.