Difference between revisions of "2025 AIME II Problems/Problem 5"
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− | Notice that <math>\triangle ABC \sim \triangle FHE \sim \triangle DEF</math> because <math>FE \parallel BC</math> and <math>DE \parallel AB,</math> so all angles in each triangle will be equal. Therefore, we have <math>\widehat{DE} = 2 \cdot 36^\circ = 72^\circ.</math> Now, quadrilateral <math>FHED</math> is cyclic, so opposite angles add to <math>180^\circ.</math> Since we know from similar triangles that <math>\angle{FED} = 60^\circ + 36^\circ = 96^\circ,</math> we see that <math>\angle{HFD} = 84^\circ.</math> We also know that <math>\angle{JFE} = 60^\circ + 36^\circ = 96^\circ,</math> so that means <math>\angle{JFH} = 96^\circ - 84^\circ = 12^\circ \implies \widehat{JH} = 24^\circ.</math> Finally, <math>\angle{B} = 60^\circ = \frac{\widehat{JED} - \widehat{FG}}{2}.</math> <math>\widehat{JED} = \widehat{DE} + \widehat{JE} = 72^\circ + 2(\angle{JFE}) = 192^\circ.</math> So <math>\widehat{FG} = 192 - 120 = 72^\circ.</math> The answer is <math>72 + 2\cdot 24 + 3\cdot 72 = \boxed{336}.</math> | + | Notice that <math>\triangle ABC \sim \triangle FHE \sim \triangle DEF</math> because <math>FE \parallel BC</math> and <math>DE \parallel AB,</math> so all angles in each triangle will be equal (this is known as the Midline Theorem). Therefore, we have <math>\widehat{DE} = 2 \cdot 36^\circ = 72^\circ.</math> Now, quadrilateral <math>FHED</math> is cyclic, so opposite angles add to <math>180^\circ.</math> Since we know from similar triangles that <math>\angle{FED} = 60^\circ + 36^\circ = 96^\circ,</math> we see that <math>\angle{HFD} = 84^\circ.</math> We also know that <math>\angle{JFE} = 60^\circ + 36^\circ = 96^\circ,</math> so that means <math>\angle{JFH} = 96^\circ - 84^\circ = 12^\circ \implies \widehat{JH} = 24^\circ.</math> Finally, <math>\angle{B} = 60^\circ = \frac{\widehat{JED} - \widehat{FG}}{2}.</math> <math>\widehat{JED} = \widehat{DE} + \widehat{JE} = 72^\circ + 2(\angle{JFE}) = 192^\circ.</math> So <math>\widehat{FG} = 192^\circ - 120^\circ = 72^\circ.</math> The answer is <math>72^\circ + 2\cdot 24^\circ + 3\cdot 72^\circ = \boxed{336^\circ}.</math> |
~[[User:grogg007|grogg007]] | ~[[User:grogg007|grogg007]] |
Latest revision as of 12:04, 26 July 2025
Contents
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.
Solution 1
Notice that due to midpoints, . As a result, the angles and arcs are readily available. Due to inscribed angles,
Similarly,
In order to calculate , we use the fact that
. We know that
, and
Substituting,
\begin{align*} 84 &= \frac{1}{2}(192-\widehat{HJ}) \\ 168 &= 192-\widehat{HJ} \\ \widehat{HJ} &= 24^\circ \end{align*}
Thus, .
~ Edited by aoum
Alternatively,
\begin{align*} \widehat{HJ} &= \widehat{FH} + \widehat{JE} - \widehat{FE} \\ &= 2\angle FEH + 2\angle JFE - 2\angle FDE \\ &= 2 \cdot 36^\circ + 2 \cdot 60^\circ - 2 \cdot 84^\circ \\ &= 24^\circ. \end{align*}
~ Pengu14
Solution 2
Notice that because
and
so all angles in each triangle will be equal (this is known as the Midline Theorem). Therefore, we have
Now, quadrilateral
is cyclic, so opposite angles add to
Since we know from similar triangles that
we see that
We also know that
so that means
Finally,
So
The answer is
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.