Difference between revisions of "1994 AIME Problems/Problem 15"
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== 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>. |
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+ | 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 of <math>P</math> (shaded region below) is simply the sum of two [[segment]]s of the circles. If we construct the midpoints of <math>M_1, M_2 = \overline{AB}, \overline{BC}</math> and note that <math>\triangle M_1BM_2 \sim \triangle ABC</math>, we see that thse segments respectively cut a <math>120^{\circ}</math> arc in the circle with radius <math>18</math> and <math>60^{\circ}</math> arc in the circle with radius <math>18\sqrt{3}</math>. | ||
+ | |||
+ | <asy> | ||
+ | pair project(pair X, pair Y, real r){return X+r*(Y-X);} | ||
+ | path endptproject(pair X, pair Y, real a, real b){return project(X,Y,a)--project(X,Y,b);} | ||
+ | |||
+ | pathpen = linewidth(1); size(250); pen dots = linetype("2 3") + linewidth(0.7), dashes = linetype("8 6")+linewidth(0.7)+blue, bluedots = linetype("1 4") + linewidth(0.7) + blue; | ||
+ | pair B = (0,0), A=(36,0), C=(0,36*3^.5), P=D(MP("P",(6,25), NE)), F = D(foot(B,A,C)); | ||
+ | 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--F,dots); D(D((B+C)/2)--F--D((A+B)/2),dots); D(C--P--B,dashes);D(P--A,dashes); | ||
+ | |||
+ | pair Fa = bisectorpoint(P,A), Fb = bisectorpoint(P,B), Fc = bisectorpoint(P,C); path La = endptproject((A+P)/2,Fa,20,-30), Lb = endptproject((B+P)/2,Fb,12,-35); D(La,bluedots);D(Lb,bluedots);D(endptproject((C+P)/2,Fc,18,-15),bluedots);D(IP(La,Lb),blue); | ||
+ | </asy> | ||
+ | |||
+ | The diagram shows <math>P</math> outside of the grayed locus; notice that the creases [the dotted blue] intersect within the triangle, which is against the problem conditions. The area of the locus is the sum of two segments of two circles; these segments cut out <math>120^{\circ}, 60^{\circ}</math> angles by simple similarity relations and angle-chasing. | ||
+ | |||
+ | 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 == | ||
{{AIME box|year=1994|num-b=14|after=Last question}} | {{AIME box|year=1994|num-b=14|after=Last question}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 21:20, 3 June 2021
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 of (shaded region below) is simply the sum of two segments of the circles. If we construct the midpoints of and note that , we see that thse segments respectively cut a arc in the circle with radius and arc in the circle with radius .
The diagram shows outside of the grayed locus; notice that the creases [the dotted blue] intersect within the triangle, which is against the problem conditions. The area of the locus is the sum of two segments of two circles; these segments cut out angles by simple similarity relations and angle-chasing.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.