Difference between revisions of "2022 SSMO Speed Round Problems/Problem 8"
(Created page with "==Problem== Circle <math>\omega</math> has chord <math>AB</math> of length <math>18</math>. Point <math>X</math> lies on chord <math>AB</math> such that <math>AX = 4.</math> C...") |
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Line 8: | Line 8: | ||
By the Pythagorean Theorem, we have | By the Pythagorean Theorem, we have | ||
− | + | <cmath>(O_1X+CO)^2+(XC)^2 = (OO_1)^2</cmath> | |
− | (O_1X+CO)^2+(XC)^2 | + | <cmath>(O_2X-CO)^2+(XC)^2 = (OO_2)^2.</cmath> |
− | (O_2X-CO)^2+(XC)^2 | ||
− | |||
which is the same as | which is the same as | ||
− | + | <cmath>(r_1+x)^2+25 = (r-r_1)^2</cmath> | |
− | (r_1+x)^2+25 = (r-r_1)^2 | + | <cmath>(r_2-x)^2+25 =(r-r_2)^2.</cmath> |
− | (r_2-x)^2+25 =(r-r_2)^2. | + | \ |
− | \ | ||
− | |||
Solving for <math>r_1</math> and <math>r_2,</math> we have that | Solving for <math>r_1</math> and <math>r_2,</math> we have that | ||
− | + | <cmath>r_1 = \frac{r^2-x^2-25}{2(r+x)}</cmath> | |
− | + | <cmath>r_2 = \frac{r^2-x^2-25}{2(r-x)}.</cmath> | |
− | |||
− | |||
Thus, | Thus, | ||
<cmath> | <cmath> | ||
Line 29: | Line 23: | ||
meaning that the minimum and maximum value of <math>r_1r_2</math> are both <math>\frac{784}{81}</math> so the answer is <math>\boxed{1649}.</math> | meaning that the minimum and maximum value of <math>r_1r_2</math> are both <math>\frac{784}{81}</math> so the answer is <math>\boxed{1649}.</math> | ||
− | + | <center> | |
− | + | <asy> | |
size(7cm); | size(7cm); | ||
point a, b, c, x, o, t, o1, o2; | point a, b, c, x, o, t, o1, o2; | ||
Line 57: | Line 51: | ||
filldraw(circle(o2, abs(o2-x)), opacity(0.2)+palered, lightred); | filldraw(circle(o2, abs(o2-x)), opacity(0.2)+palered, lightred); | ||
− | dot(" | + | dot("$A$", a, dir(145)); |
− | dot(" | + | dot("$B$", b, dir(30)); |
− | dot(" | + | dot("$C$", c, dir(90)); |
− | dot(" | + | dot("$X$", x, dir(60)); |
− | dot(" | + | dot("$O$", o, dir(45)); |
− | dot(" | + | dot("$O_1$", o1); |
− | dot(" | + | dot("$O_2$", o2); |
− | + | </asy> | |
− | + | </center> |
Revision as of 13:03, 3 July 2023
Problem
Circle has chord
of length
. Point
lies on chord
such that
Circle
with radius
and
with radius
lie on two different sides of
Both
and
are tangent to
at
and
If the sum of the maximum and minimum values of
is
find
.
Solution
Let be the radius of
and let
be the midpoint of
and let
Note that
. WLOG assume that
Since and
we have
By the Pythagorean Theorem, we have
which is the same as
\
Solving for
and
we have that
Thus,
meaning that the minimum and maximum value of
are both
so the answer is
size(7cm); point a, b, c, x, o, t, o1, o2; a = (0,0); b = (18,0); c = (9,0); x = (4,0); o = (9, -3); circle cir = circle(o, abs(a-o)); t = intersectionpoints(cir, line(x,o))[1]; point[] o1o2 = intersectionpoints(ellipse(x, o, (x+t)/2), line(x, x+(0,1))); o1 = o1o2[0]; o2 = o1o2[1]; draw(o1--o2, red); draw(a--b, blue); draw(c--o, blue); filldraw(cir, opacity(0.2)+lightcyan, blue); // draw(ellipse(x, o, (x+t)/2)); filldraw(circle(o1, abs(o1-x)), opacity(0.2)+palered, lightred); filldraw(circle(o2, abs(o2-x)), opacity(0.2)+palered, lightred); dot("$A$", a, dir(145)); dot("$B$", b, dir(30)); dot("$C$", c, dir(90)); dot("$X$", x, dir(60)); dot("$O$", o, dir(45)); dot("$O_1$", o1); dot("$O_2$", o2); (Error making remote request. Unknown error_msg)