Difference between revisions of "1983 AIME Problems/Problem 12"
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== Problem == | == Problem == | ||
− | + | Diameter <math>AB</math> of a circle has length a <math>2</math>-digit integer (base ten). Reversing the digits gives the length of the perpendicular chord <math>CD</math>. The distance from their intersection point <math>H</math> to the center <math>O</math> is a positive rational number. Determine the length of <math>AB</math>. | |
<asy> | <asy> | ||
− | + | draw(circle((0,0),4)); | |
− | + | draw((-4,0)--(4,0)); | |
− | + | draw((-2,-2*sqrt(3))--(-2,2*sqrt(3))); | |
− | + | draw((-2.6,0)--(-2.6,0.6)); | |
− | + | draw((-2,0.6)--(-2.6,0.6)); | |
− | </asy | + | dot((0,0)); |
+ | dot((-2,0)); | ||
+ | dot((4,0)); | ||
+ | dot((-4,0)); | ||
+ | dot((-2,2*sqrt(3))); | ||
+ | dot((-2,-2*sqrt(3))); | ||
+ | label("A",(-4,0),W); | ||
+ | label("B",(4,0),E); | ||
+ | label("C",(-2,2*sqrt(3)),NW); | ||
+ | label("D",(-2,-2*sqrt(3)),SW); | ||
+ | label("H",(-2,0),SE); | ||
+ | label("O",(0,0),NE);</asy> | ||
== Solution == | == Solution == | ||
− | Let <math>AB=10x+y</math> and <math>CD=10y+x</math>. It follows that <math>CO=\frac{AB}{2}=\frac{10x+y}{2}</math> and <math>CH=\frac{CD}{2}=\frac{10y+x}{2}</math>. Applying the [[Pythagorean Theorem]] on <math> | + | Let <math>AB=10x+y</math> and <math>CD=10y+x</math>. It follows that <math>CO=\frac{AB}{2}=\frac{10x+y}{2}</math> and <math>CH=\frac{CD}{2}=\frac{10y+x}{2}</math>. Scale up this triangle by 2 to ease the arithmetic. Applying the [[Pythagorean Theorem]] on <math>2CO</math>, <math>2OH</math> and <math>2CH</math>, we deduce |
+ | <cmath>(2OH)^2=\left(10x+y\right)^2-\left(10y+x\right)^2=99(x+y)(x-y)</cmath> | ||
+ | |||
+ | Because <math>OH</math> is a positive rational number and <math>x</math> and <math>y</math> are integral, the quantity <math>99(x+y)(x-y)</math> must be a perfect square. Hence either <math>x-y</math> or <math>x+y</math> must be a multiple of <math>11</math>, but as <math>x</math> and <math>y</math> are different digits, <math>1+0=1 \leq x+y \leq 9+9=18</math>, so the only possible multiple of <math>11</math> is <math>11</math> itself. However, <math>x-y</math> cannot be 11, because both must be digits. Therefore, <math>x+y</math> must equal <math>11</math> and <math>x-y</math> must be a perfect square. The only pair <math>(x,y)</math> that satisfies this condition is <math>(6,5)</math>, so our answer is <math>\boxed{065}</math>. (Therefore <math>CD = 56</math> and <math>OH = \frac{33}{2}</math>.) | ||
+ | |||
+ | == Alternate start to solution 1 == | ||
+ | Since H is the midpoint of <math>CD</math>, by [[Power of a Point]], <math>CH^2=(AH)(BH)</math>. Because <math>AH=r-OH</math> and <math>BH=r+OH</math>, where <math>r</math> is the radius of the circle, we deduce the relation <math>CH^2=r^2-OH^2</math>. Thus <math>OH^2=\left(\frac{10x+y}{2}\right)^2-\left(\frac{10y+x}{2}\right)^2</math>. Continue as above. | ||
− | + | ~anduran | |
− | == See | + | == See Also == |
{{AIME box|year=1983|num-b=11|num-a=13}} | {{AIME box|year=1983|num-b=11|num-a=13}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 17:23, 1 January 2024
Problem
Diameter of a circle has length a -digit integer (base ten). Reversing the digits gives the length of the perpendicular chord . The distance from their intersection point to the center is a positive rational number. Determine the length of .
Solution
Let and . It follows that and . Scale up this triangle by 2 to ease the arithmetic. Applying the Pythagorean Theorem on , and , we deduce
Because is a positive rational number and and are integral, the quantity must be a perfect square. Hence either or must be a multiple of , but as and are different digits, , so the only possible multiple of is itself. However, cannot be 11, because both must be digits. Therefore, must equal and must be a perfect square. The only pair that satisfies this condition is , so our answer is . (Therefore and .)
Alternate start to solution 1
Since H is the midpoint of , by Power of a Point, . Because and , where is the radius of the circle, we deduce the relation . Thus . Continue as above.
~anduran
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |