Difference between revisions of "1994 AIME Problems/Problem 15"
(category) |
(solution) |
||
Line 3: | Line 3: | ||
== Solution == | == Solution == | ||
− | {{ | + | Let <math>O_{AB}</math> be the intersection of the [[perpendicular bisector]]s (in other words, the intersections of the creases) of <math>\overline{PA}</math> and <math>\overline{PB}</math>, and so forth. Then <math>O_{AB}, O_{BC}, O_{CA}</math> are, respectively, the [[circumcenter]]s of <math>\triangle PAB, PBC, PCA</math>. According to the problem statement, the circumcenters of the triangles cannot lie within the interior of the respective triangles, since they are not on the paper. It follows that <math>\angle APB, \angle BPC, \angle CPA > 90^{\circ}</math>; the locus of each of the respective conditions for <math>P</math> is the region inside the (semi)circles with diameters <math>\overline{AB}, \overline{BC}, \overline{CA}</math>. |
+ | |||
+ | We note that the circle with diameter <math>AC</math> covers the entire triangle because it is the circumcircle of <math>\triangle ABC</math>, so it suffices to take the intersection of the circles about <math>AB, BC</math>. We note that their intersection lies entirely within <math>\triangle ABC</math> (the chord connecting the endpoints of the region is in fact the altitude of <math>\triangle ABC</math> from <math>B</math>). Thus, the area of the locus is simply the sum of two segments of the circles, respectively cutting a <math>120^{\circ}</math> arc in the circle with radius <math>18</math> and <math>60^{\circ}</math> in the circle with radius <math>18\sqrt{3}</math>. | ||
+ | |||
+ | <center><asy> | ||
+ | pathpen = linewidth(0.7); size(200); pen dots = linetype("2 2") + linewidth(0.7); | ||
+ | pair B = (0,0), A=(36,0), C=(0,36*3^.5), P=D(MP("P",(10,25), NE)); | ||
+ | D(D(MP("A",A)) -- D(MP("B",B)) -- D(MP("C",C,N)) -- cycle); | ||
+ | fill(arc((A+B)/2,18,60,180) -- arc((B+C)/2,18*3^.5,-90,-30) -- cycle, rgb(0.8,0.8,0.8)); D(arc((A+B)/2,18,0,180),dots); D(arc((B+C)/2,18*3^.5,-90,90),dots); D(arc((A+C)/2,36,120,300),dots); D(B--foot(B,A,C),dots); D(C--P--B--P--A,linetype("6 6")+linewidth(0.7)); | ||
+ | </asy></center> | ||
+ | |||
+ | Hence, the answer is, using the <math>\frac 12 ab\sin C</math> definition of triangle area, <math>\left[\frac{\pi}{3} \cdot 18^2 - \frac{1}{2} \cdot 18^2 \sin \frac{2\pi}{3} \right] + \left[\frac{\pi}{6} \cdot \left(18\sqrt{3}\right)^2 - \frac{1}{2} \cdot (18\sqrt{3})^2 \sin \frac{\pi}{3}\right] = 270\pi - 324\sqrt{3}</math>, and <math>q+r+s = \boxed{597}</math>. | ||
== See also == | == See also == |
Revision as of 16:55, 9 December 2008
Problem
Given a point on a triangular piece of paper
consider the creases that are formed in the paper when
and
are folded onto
Let us call
a fold point of
if these creases, which number three unless
is one of the vertices, do not intersect. Suppose that
and
Then the area of the set of all fold points of
can be written in the form
where
and
are positive integers and
is not divisible by the square of any prime. What is
?
Solution
Let be the intersection of the perpendicular bisectors (in other words, the intersections of the creases) of
and
, and so forth. Then
are, respectively, the circumcenters of
. According to the problem statement, the circumcenters of the triangles cannot lie within the interior of the respective triangles, since they are not on the paper. It follows that
; the locus of each of the respective conditions for
is the region inside the (semi)circles with diameters
.
We note that the circle with diameter covers the entire triangle because it is the circumcircle of
, so it suffices to take the intersection of the circles about
. We note that their intersection lies entirely within
(the chord connecting the endpoints of the region is in fact the altitude of
from
). Thus, the area of the locus is simply the sum of two segments of the circles, respectively cutting a
arc in the circle with radius
and
in the circle with radius
.
![[asy] pathpen = linewidth(0.7); size(200); pen dots = linetype("2 2") + linewidth(0.7); pair B = (0,0), A=(36,0), C=(0,36*3^.5), P=D(MP("P",(10,25), NE)); D(D(MP("A",A)) -- D(MP("B",B)) -- D(MP("C",C,N)) -- cycle); fill(arc((A+B)/2,18,60,180) -- arc((B+C)/2,18*3^.5,-90,-30) -- cycle, rgb(0.8,0.8,0.8)); D(arc((A+B)/2,18,0,180),dots); D(arc((B+C)/2,18*3^.5,-90,90),dots); D(arc((A+C)/2,36,120,300),dots); D(B--foot(B,A,C),dots); D(C--P--B--P--A,linetype("6 6")+linewidth(0.7)); [/asy]](http://latex.artofproblemsolving.com/c/e/4/ce4a76c5965d6e532ffa19a9434a7814d15a7bff.png)
Hence, the answer is, using the definition of triangle area,
, and
.
See also
1994 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |