Difference between revisions of "2025 AIME II Problems/Problem 5"
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<cmath>\widehat{FG}=2\angle FDB=2\angle ACB=2\cdot36=72^\circ</cmath> | <cmath>\widehat{FG}=2\angle FDB=2\angle ACB=2\cdot36=72^\circ</cmath> | ||
| − | In order to calculate <math>\widehat{HJ}</math>, we use the fact that <math>\angle BAC=\frac{1}{2}(\widehat{FDE}-\widehat{HJ})</math>. We know that | + | In order to calculate <math>\widehat{HJ}</math>, we use the fact that <math>\angle BAC=\frac{1}{2}(\widehat{FDE}-\widehat{HJ})</math>. We know that <math>\angle BAC=84^\circ</math>, and |
| − | < | + | <cmath>\widehat{FDE}=360-\widehat{FE}=360-2\angle FDE=360-2\angle CAB=360-2\cdot84=192^\circ</cmath> |
Substituting, | Substituting, | ||
| − | < | + | <cmath>84=\frac{1}{2}(192-\widehat{HJ})</cmath> |
| − | < | + | <cmath>168=192-\widehat{HJ}</cmath> |
| − | < | + | <cmath>\widehat{HJ}=24</cmath> |
Thus, <math>\widehat{DE}+2\cdot\widehat{HJ}+3\cdot\widehat{FG}=72+48+216=\boxed{336}^\circ</math>. ~eevee9406 | Thus, <math>\widehat{DE}+2\cdot\widehat{HJ}+3\cdot\widehat{FG}=72+48+216=\boxed{336}^\circ</math>. ~eevee9406 | ||
Revision as of 02:19, 14 February 2025
Problem
Suppose 
has angles 
and 
Let 
and 
be the midpoints of sides 
and 
respectively. The circumcircle of 
intersects 
and 
at points 
and 
respectively. The points 
and divide the circumcircle of into six minor arcs, as shown. Find 
where the arcs are measured in degrees.
![[asy] import olympiad; size(6cm); defaultpen(fontsize(10pt)); pair B = (0, 0), A = (Cos(60), Sin(60)), C = (Cos(60)+Sin(60)/Tan(36), 0), D = midpoint(B--C), E = midpoint(A--C), F = midpoint(A--B); guide circ = circumcircle(D, E, F); pair G = intersectionpoint(B--D, circ), J = intersectionpoints(A--F, circ)[0], H = intersectionpoints(A--E, circ)[0]; draw(B--A--C--cycle); draw(D--E--F--cycle); draw(circ); dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(G);dot(H);dot(J); label("$A$", A, (0, .8)); label("$B$", B, (-.8, -.8)); label("$C$", C, (.8, -.8)); label("$D$", D, (0, -.8)); label("$E$", E, (.8, .2)); label("$F$", F, (-.8, .2)); label("$G$", G, (0, .8)); label("$H$", H, (-.2, -1)); label("$J$", J, (.2, -.8)); [/asy]](http://latex.artofproblemsolving.com/8/9/2/8927c2fea7d62dafca511d88a32813611716ed6f.png)
Solution
Notice that due to midpoints, 
. As a result, the angles and arcs are readily available. Due to inscribed angles,
![\[\widehat{DE}=2\angle DFE=2\angle ACB=2\cdot36=72^\circ\]](http://latex.artofproblemsolving.com/9/f/2/9f213ee2a8b7976a556313cf62cd2d264c07328c.png)
Similarly,
![\[\widehat{FG}=2\angle FDB=2\angle ACB=2\cdot36=72^\circ\]](http://latex.artofproblemsolving.com/2/4/6/246acc906b28733766a4cf0a24929591408837b6.png)
In order to calculate 
, we use the fact that 
. We know that 
, and
![\[\widehat{FDE}=360-\widehat{FE}=360-2\angle FDE=360-2\angle CAB=360-2\cdot84=192^\circ\]](http://latex.artofproblemsolving.com/c/d/d/cdde1515fef7ddcf2f524280a7abf2218bc08a83.png)
Substituting,
![\[84=\frac{1}{2}(192-\widehat{HJ})\]](http://latex.artofproblemsolving.com/2/a/5/2a5dadf5613943bcedb493eb4683dd0865a92c3c.png)
![\[168=192-\widehat{HJ}\]](http://latex.artofproblemsolving.com/2/f/e/2fe9c6f304c752ac5ab3a0b29fcac093128b528d.png)
![\[\widehat{HJ}=24\]](http://latex.artofproblemsolving.com/e/b/9/eb9ac73fbf9ef3c8ee54f09e3bf37e3e2fed35ff.png)
Thus, 
. ~eevee9406
See also
| 2025 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
