1983 AIME Problems/Problem 4
Contents
Problem
A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is cm, the length of
is
cm and that of
is
cm. The angle
is a right angle. Find the square of the distance (in centimeters) from
to the center of the circle.
Solution 1
Because we are given a right angle, we look for ways to apply the Pythagorean Theorem. Let the foot of the perpendicular from to
be
and let the foot of the perpendicular from
to the line
be
. Let
and
. We're trying to find
.
![[asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r); pair D=(A.x,0),F=(0,B.y); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); draw(P); draw(C--B--O--A--B); draw(D--O--F--B,dashed); dot(O); dot(A); dot(B); dot(C); label("$O$",O,SW); label("$A$",A,NE); label("$B$",B,S); label("$C$",C,SE); label("$D$",D,NE); label("$E$",F,SW); [/asy]](http://latex.artofproblemsolving.com/b/c/a/bcac965883252498b864ad906b153cd402838000.png)
Applying the Pythagorean Theorem, and
.
Thus, , and
. We solve this system to get
and
, such that the answer is
.
Solution 2
We'll use the law of cosines. Let be the center of the circle; we wish to find
. We know how long
and
are, so if we can find
, we'll be in good shape.
![[asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r); pair D=(A.x,0),F=(0,B.y); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); draw(P); draw(C--B--O--A--B); draw(O--B); draw(A--C); dot(O); dot(A); dot(B); dot(C); label("$O$",O,SW); label("$A$",A,NE); label("$B$",B,S); label("$C$",C,SE); [/asy]](http://latex.artofproblemsolving.com/5/5/7/557f2e442b02c5c1ad40cbfdd66bc18626030627.png)
We can find using angles
and
. First we note that by Pythagoras,
If we let
be the midpoint of
, that mean that
. Since
is isosceles (
from the definition of a circle),
is also the foot of the altitude from
to
![[asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r); pair D=(A.x,0); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); pair M = (A+C)/2; draw(P); draw(O--C--A--cycle); draw(O--M, dashed); draw(rightanglemark(O,M,A,25)); dot(O); dot(A); dot(C); label("$O$",O,SW); label("$A$",A,NE); label("$M$",M,SSW); label("$C$",C,SE); label("$\sqrt{50}$", (O+A)/2, NW); label("$\sqrt{10}$", (A+M)/2, E); [/asy]](http://latex.artofproblemsolving.com/8/0/4/804ba7a0599d3ff45160b45a45e6819be9e2aea2.png)
It follows that . Therefore
Meanwhile, from right triangle
we have
This means that by the angle subtraction formulas,
Now we have all we need to use the law of cosines on This tells us that
Solution 3
Drop perpendiculars from to
(with foot
),
to
(with foot
), and
to
(with foot
).
Also, mark the midpoint
of
.
![[asy] size(200); pair dl(string name, pair loc, pair offset) { dot(loc); label(name,loc,offset); return loc; }; pair a[] = {(0,0),(0,5),(1,5),(1,7),(-2,6),(-5,5),(-2,5),(-2,6),(0,6)}; string n[] = {"O","$T_1$","B","C","M","A","$T_3$","M","$T_2$"}; for(int i=0;i<a.length;++i) { dl(n[i],a[i],dir(degrees(a[i],false) ) ); draw(a[(i-1)%a.length]--a[i]); }; dot(a); draw(a[5]--a[1]); draw(a[0]--a[3]); draw(a[0]--a[4]); draw(a[0]--a[2]); draw(a[0]--a[5]); draw(a[5]--a[2]--a[3]--cycle,blue+linewidth(0.7)); draw(a[0]--a[8]--a[7]--cycle,blue+linewidth(0.7)); [/asy]](http://latex.artofproblemsolving.com/e/c/8/ec8178dd6f1c58944b30643b6498df4a33724323.png)
First notice that by computation, is a
isosceles triangle, so
.
Then, notice that
. Therefore, the two blue triangles are congruent, from which we deduce
and
. As
and
, we subtract and get
. Then the Pythagorean Theorem tells us that
.
Solution 4
Draw segment with length
, and draw radius
such that
bisects chord
at point
. This also means that
is perpendicular to
. By the Pythagorean Theorem, we get that
, and therefore
. Also by the Pythagorean theorem, we can find that
.
Next, find and
. Since
, we get
By the subtraction formula for
, we get
Finally, by the Law of Cosines on
, we get
Solution 5
We use coordinates. Let the circle have center and radius
; this circle has equation
. Let the coordinates of
be
. We want to find
.
and
with coordinates
and
, respectively, both lie on the circle. From this we obtain the system of equations
After expanding these terms, we notice by subtracting the first and second equations, we can cancel out and
. after substituting
and plugging back in, we realize that
or
. Since the first point is out of the circle, we find that
is the only relevant answer. This paragraph is written by ~hastapasta.
Solving, we get and
, so the distance is
.
Solution 6 (Trigonometry)
![[asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r); pair D=(A.x,0),F=(0,B.y); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); draw(P); draw(C--B--O--A--B); draw(O--B); draw(A--C); draw(O--C); dot(O); dot(A); dot(B); dot(C); label("$O$",O,SW); label("$A$",A,NE); label("$B$",B,S); label("$C$",C,SE); [/asy]](http://latex.artofproblemsolving.com/2/1/4/21489b007ad01ff0bc01a413040695a075e5d499.png)
I will use the law of cosines in triangle and
.
Video Solution by OmegaLearn
https://youtu.be/nPVDavMoG9M?t=353
~ pi_is_3.14
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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