Difference between revisions of "1983 AIME Problems/Problem 4"
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− | == Solution == | + | == Solutions == |
+ | === 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 <math>O</math> to <math>AB</math> be <math>D</math> and let the foot of the perpendicular from <math>O</math> to the [[line]] <math>BC</math> be <math>E</math>. Let <math>OE=x</math> and <math>OD=y</math>. We're trying to find <math>x^2+y^2</math>. | Because we are given a right angle, we look for ways to apply the [[Pythagorean Theorem]]. Let the foot of the [[perpendicular]] from <math>O</math> to <math>AB</math> be <math>D</math> and let the foot of the perpendicular from <math>O</math> to the [[line]] <math>BC</math> be <math>E</math>. Let <math>OE=x</math> and <math>OD=y</math>. We're trying to find <math>x^2+y^2</math>. | ||
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Thus, <math>(\sqrt{50})^2 = y^2 + (6-x)^2</math>, and <math>(\sqrt{50})^2 = x^2 + (y+2)^2</math>. We solve this system to get <math>x = 1</math> and <math>y = 5</math>, resulting in an answer of <math>1^2 + 5^2 = \boxed{026}</math>. | Thus, <math>(\sqrt{50})^2 = y^2 + (6-x)^2</math>, and <math>(\sqrt{50})^2 = x^2 + (y+2)^2</math>. We solve this system to get <math>x = 1</math> and <math>y = 5</math>, resulting in an answer of <math>1^2 + 5^2 = \boxed{026}</math>. | ||
− | == Synthetic Solution == | + | === Solution 2: Synthetic Solution === |
Drop perpendiculars from <math>O</math> to <math>AB</math> (<math>T_1</math>), <math>M</math> to <math>OT_1</math> (<math>T_2</math>), and <math>M</math> to <math>AB</math> (<math>T_3</math>). | Drop perpendiculars from <math>O</math> to <math>AB</math> (<math>T_1</math>), <math>M</math> to <math>OT_1</math> (<math>T_2</math>), and <math>M</math> to <math>AB</math> (<math>T_3</math>). | ||
Also, draw the midpoint <math>M</math> of <math>AC</math>. | Also, draw the midpoint <math>M</math> of <math>AC</math>. |
Revision as of 08:50, 30 September 2008
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)
Solutions
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 an answer of
.
Solution 2: Synthetic Solution
Drop perpendiculars from to
(
),
to
(
), and
to
(
).
Also, draw the midpoint
of
.
Then the problem is trivialized. Why?
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
.
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 |