Difference between revisions of "1997 AJHSME Problems/Problem 12"
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draw(H--I--J--cycle); | draw(H--I--J--cycle); | ||
draw(K--L--J); | draw(K--L--J); | ||
| − | draw(arc((0,0), | + | draw(arc((0,0),,(1,0),CW)); label("$70^\circ$",dir(35),NE); |
draw(arc(I,I+dir(250),I+dir(290),CCW)); label("$40^\circ$",I+1.25*dir(270),S); | draw(arc(I,I+dir(250),I+dir(290),CCW)); label("$40^\circ$",I+1.25*dir(270),S); | ||
label("$1$",J+0.25*dir(162.5),NW); label("$2$",J+0.25*dir(17.5),NE); | label("$1$",J+0.25*dir(162.5),NW); label("$2$",J+0.25*dir(17.5),NE); | ||
Revision as of 19:09, 23 April 2021
Problem
Find 
pair H,I,J,K,L;
H = (0,0); I = 10*dir(70); J = I + 10*dir(290); K = J + 5*dir(110); L = J + 5*dir(0);
draw(H--I--J--cycle);
draw(K--L--J);
draw(arc((0,0),,(1,0),CW)); label("$70^\circ$",dir(35),NE);
draw(arc(I,I+dir(250),I+dir(290),CCW)); label("$40^\circ$",I+1.25*dir(270),S);
label("$1$",J+0.25*dir(162.5),NW); label("$2$",J+0.25*dir(17.5),NE);
label("$3$",L+dir(162.5),WNW); label("$4$",K+dir(-52.5),SE);
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Solution
Using the left triangle, we have:
Using the given fact that 
, we have 
.
Finally, using the right triangle, and the fact that , we have:
Thus, the answer is 
See also
| 1997 AJHSME (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 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
