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Difference between revisions of "1994 AIME Problems/Problem 15"

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== Solution ==
 
== Solution ==
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{{solution}}
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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>.
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<asy>
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pair project(pair X, pair Y, real r){return X+r*(Y-X);}
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path endptproject(pair X, pair Y, real a, real b){return project(X,Y,a)--project(X,Y,b);}
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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;
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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));
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D(D(MP("A",A)) -- D(MP("B",B)) -- D(MP("C",C,N)) -- cycle);
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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);
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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);
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</asy>
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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.
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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}}
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[[Category:Intermediate Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 21:20, 3 June 2021

Problem

Given a point $P^{}_{}$ on a triangular piece of paper $ABC,\,$ consider the creases that are formed in the paper when $A, B,\,$ and $C\,$ are folded onto $P.\,$ Let us call $P_{}^{}$ a fold point of $\triangle ABC\,$ if these creases, which number three unless $P^{}_{}$ is one of the vertices, do not intersect. Suppose that $AB=36, AC=72,\,$ and $\angle B=90^\circ.\,$ Then the area of the set of all fold points of $\triangle ABC\,$ can be written in the form $q\pi-r\sqrt{s},\,$ where $q, r,\,$ and $s\,$ are positive integers and $s\,$ is not divisible by the square of any prime. What is $q+r+s\,$?

Solution

Let $O_{AB}$ be the intersection of the perpendicular bisectors (in other words, the intersections of the creases) of $\overline{PA}$ and $\overline{PB}$, and so forth. Then $O_{AB}, O_{BC}, O_{CA}$ are, respectively, the circumcenters of $\triangle PAB, PBC, PCA$. 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 $\angle APB, \angle BPC, \angle CPA > 90^{\circ}$; the locus of each of the respective conditions for $P$ is the region inside the (semi)circles with diameters $\overline{AB}, \overline{BC}, \overline{CA}$.

We note that the circle with diameter $AC$ covers the entire triangle because it is the circumcircle of $\triangle ABC$, so it suffices to take the intersection of the circles about $AB, BC$. We note that their intersection lies entirely within $\triangle ABC$ (the chord connecting the endpoints of the region is in fact the altitude of $\triangle ABC$ from $B$). Thus, the area of the locus of $P$ (shaded region below) is simply the sum of two segments of the circles. If we construct the midpoints of $M_1, M_2 = \overline{AB}, \overline{BC}$ and note that $\triangle M_1BM_2 \sim \triangle ABC$, we see that thse segments respectively cut a $120^{\circ}$ arc in the circle with radius $18$ and $60^{\circ}$ arc in the circle with radius $18\sqrt{3}$.

[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 $P$ 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 $120^{\circ}, 60^{\circ}$ angles by simple similarity relations and angle-chasing.

Hence, the answer is, using the $\frac 12 ab\sin C$ definition of triangle area, $\left[\frac{\pi}{3} \cdot 18^2 - \frac{1}{2} \cdot 18^2 \sin \frac{2\pi}{3} \right] + \left[\frac{\pi}{6} \cdot \left(18\sqrt{3}\right)^2 - \frac{1}{2} \cdot (18\sqrt{3})^2 \sin \frac{\pi}{3}\right] = 270\pi - 324\sqrt{3}$, and $q+r+s = \boxed{597}$.

See also

1994 AIME (Problems • Answer Key • Resources)
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