Difference between revisions of "2015 AMC 10B Problems/Problem 19"
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draw((3.46,0.96)--(8.02,-3.44)--(12.42,1.12)--(7.86,5.52)--cycle); | draw((3.46,0.96)--(8.02,-3.44)--(12.42,1.12)--(7.86,5.52)--cycle); | ||
/* draw figures */ | /* draw figures */ | ||
| + | draw((3.34,-3.86)--(3.14,-3.66)--(3.34,-3.66)) | ||
draw((3.46,0.96)--(3.44,-3.36)); | draw((3.46,0.96)--(3.44,-3.36)); | ||
draw((3.44,-3.36)--(8.02,-3.44)); | draw((3.44,-3.36)--(8.02,-3.44)); | ||
| Line 51: | Line 52: | ||
label("$O$", (5.52,-1.82), NE * labelscalefactor); | label("$O$", (5.52,-1.82), NE * labelscalefactor); | ||
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
| − | |||
</asy> | </asy> | ||
Revision as of 13:11, 26 October 2018
Problem
In 
, 
and 
. Squares 
and 
are constructed outside of the triangle. The points 
, and 
lie on a circle. What is the perimeter of the triangle?
Solution
The center of the circle lies on the perpendicular bisectors of both chords 
and 
. Therefore we know the center of the circle must also be the midpoint of the hypotenuse. Let this point be 
. Draw perpendiculars to and from , and connect 
and 
. 
. Let 
and 
. Then 
. Simplifying this gives 
. But by Pythagorean Theorem on 
, we know 
, because 
. Thus 
. So our equation simplifies further to 
. However , so 
, which means 
, or 
. Aha! This means is just an isosceles right triangle, so 
, and thus the perimeter is 
.
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(11.5cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -4.3, xmax = 18.7, ymin = -5.26, ymax = 6.3; /* image dimensions */
draw((3.46,0.96)--(3.44,-3.36)--(8.02,-3.44)--cycle);
draw((3.46,0.96)--(8.02,-3.44)--(12.42,1.12)--(7.86,5.52)--cycle);
/* draw figures */
draw((3.34,-3.86)--(3.14,-3.66)--(3.34,-3.66))
draw((3.46,0.96)--(3.44,-3.36));
draw((3.44,-3.36)--(8.02,-3.44));
draw((8.02,-3.44)--(3.46,0.96));
draw((3.46,0.96)--(-0.86,0.98));
draw((-0.86,0.98)--(-0.88,-3.34));
draw((-0.88,-3.34)--(3.44,-3.36));
draw((3.46,0.96)--(8.02,-3.44));
draw((8.02,-3.44)--(12.42,1.12));
draw((12.42,1.12)--(7.86,5.52));
draw((7.86,5.52)--(3.46,0.96));
draw((5.74,-1.24)--(-0.86,0.98));
draw((5.74,-1.24)--(-0.87,-1.18), linetype("4 4"));
draw((5.74,-1.24)--(7.86,5.52));
draw((5.74,-1.24)--(10.14,3.32), linetype("4 4"));
draw(shift((5.82,-1.21))*xscale(6.99920709795045)*yscale(6.99920709795045)*arc((0,0),1,19.44457562540183,197.63600413408128), linetype("2 2"));
/* dots and labels */
dot((3.46,0.96),dotstyle);
label("$A$", (3.2,1.06), NE * labelscalefactor);
dot((3.44,-3.36),dotstyle);
label("$C$", (3.14,-3.86), NE * labelscalefactor);
dot((8.02,-3.44),dotstyle);
label("$B$", (8.06,-3.8), NE * labelscalefactor);
dot((-0.86,0.98),dotstyle);
label("$Z$", (-1.34,1.12), NE * labelscalefactor);
dot((-0.88,-3.34),dotstyle);
label("$W$", (-1.48,-3.54), NE * labelscalefactor);
dot((12.42,1.12),dotstyle);
label("$X$", (12.5,1.24), NE * labelscalefactor);
dot((7.86,5.52),dotstyle);
label("$Y$", (7.94,5.64), NE * labelscalefactor);
dot((5.74,-1.24),dotstyle);
label("$O$", (5.52,-1.82), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
(Error making remote request. Unknown error_msg)
See Also
| 2015 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
