Difference between revisions of "1983 AIME Problems/Problem 4"
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So, <math>MT_2 = 2,OT_2 = 6</math>. As <math>T_3B = 3, MT_3 = 1</math>, we subtract and get <math>OT_1 = 5,T_1B = 1</math>. Then the Pythagorean Theorem shows <math>OB^2 = \boxed{026}</math>. | So, <math>MT_2 = 2,OT_2 = 6</math>. As <math>T_3B = 3, MT_3 = 1</math>, we subtract and get <math>OT_1 = 5,T_1B = 1</math>. Then the Pythagorean Theorem shows <math>OB^2 = \boxed{026}</math>. | ||
+ | |||
+ | === Solution 3 === | ||
+ | Draw segment <math>OB</math> with length <math>x</math> and radius <math>OQ</math> such that <math>OQ</math> bisects chord <math>AC</math> at point <math>M</math>. This also means that <math>OQ</math> is perpendicular to <math>AC</math>. By the Pythagorean Theorem, we get that <math>AC=\sqrt{(BC)^2+(AB)^2}=2\sqrt{10}</math>, and therefore <math>AM=\sqrt{10}</math>. Also by the Pythagorean theorem, we can find that <math>OM=\sqrt{50-10}=2\sqrt{10}</math>. | ||
+ | |||
+ | Next, find <math>\angle BAC=\arctan{(\frac{2}{6})}</math> and <math>\angle OAM=\arctan{(\frac{2\sqrt{10}}{\sqrt{10}})}</math>. Since <math>\angle OAB=\angle OAM-\angle BAC</math>, we get <cmath>\angle OAB=\arctan{2}-\arctan{\frac{1}{3}}</cmath><cmath>\tan{(\angle OAB)}=\tan{(\arctan{2}-\arctan{\frac{1}{3}})}</cmath>By the tangent angle subtraction formula, we get<cmath>\tan{(\angle OAB)}=\frac{2-\frac{1}{3}}{1+2\cdot \frac{1}{3}}</cmath><cmath>\tan{(\angle OAB)}=1</cmath><cmath>\cos{(\angle OAB)}=\frac{1}{\sqrt{2}}</cmath>Finally, by the Law of Cosines on <math>\triangle OAB</math>, we get <cmath>x^2=50+36-2(6)\sqrt{50}\frac{1}{\sqrt{2}}</cmath><cmath>x^2=\boxed{026}</cmath> | ||
== See Also == | == See Also == |
Revision as of 21:37, 12 June 2018
Problem
A machine shop cutting tool is in the shape of a notched circle, as shown. The radius of the circle is cm, the length of
is 6 cm, and that of
is 2 cm. The angle
is a right angle. Find the square of the distance (in centimeters) from
to the center of the circle.
![[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),C; path P=circle(O,r); C=intersectionpoint(B--(B.x+r,B.y),P); draw(P); draw(C--B--O--A--B); 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/1/7/6/176e7aff44ffcb4a35dde174c0fde108ce69cbe7.png)
Solution
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),C; pair D=(A.x,0),F=(0,B.y); path P=circle(O,r); 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/0/0/5/005734c82fcbb09bb1717995c009ee75839e265e.png)
Applying the Pythagorean Theorem, and
.
Thus, , and
. We solve this system to get
and
, resulting in
.
Solution 2
Drop perpendiculars from to
(
),
to
(
), and
to
(
).
Also, draw the midpoint
of
.
Then the problem is trivialized. Why?
![[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; thus
.
Then, notice that
. Thus the two blue triangles are congruent.
So, . As
, we subtract and get
. Then the Pythagorean Theorem shows
.
Solution 3
Draw segment with length
and 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 tangent angle subtraction formula, we get
Finally, by the Law of Cosines on
, we get
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 | ||
All AIME Problems and Solutions |