Difference between revisions of "2025 AIME II Problems/Problem 5"
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== Problem == | == Problem == | ||
− | Suppose <math>\triangle ABC</math> has angles <math>\angle BAC = 84^\circ, \angle ABC=60^\circ,</math> and <math>\angle ACB = 36^\circ.</math> Let <math>D, E,</math> and <math>F</math> be the midpoints of sides <math>\overline{BC}, \overline{AC},</math> and <math>\overline{AB},</math> respectively. The circumcircle of <math>\triangle DEF</math> intersects <math>\overline{BD}, \overline{AE},</math> and <math>\overline{AF}</math> at points <math>G, H,</math> and <math>J,</math> respectively. The points <math>G, D, E, H, J,</math> and <math>F</math> divide the circumcircle of <math>\triangle DEF</math> into six minor arcs, as shown. Find <math>\widehat{DE}+2\cdot \widehat{HJ} + 3\widehat{FG},</math> where the arcs are measured in degrees. | + | Suppose <math>\triangle ABC</math> has angles <math>\angle BAC = 84^\circ, \angle ABC=60^\circ,</math> and <math>\angle ACB = 36^\circ.</math> Let <math>D, E,</math> and <math>F</math> be the midpoints of sides <math>\overline{BC}, \overline{AC},</math> and <math>\overline{AB},</math> respectively. The circumcircle of <math>\triangle DEF</math> intersects <math>\overline{BD}, \overline{AE},</math> and <math>\overline{AF}</math> at points <math>G, H,</math> and <math>J,</math> respectively. The points <math>G, D, E, H, J,</math> and <math>F</math> divide the circumcircle of <math>\triangle DEF</math> into six minor arcs, as shown. Find <math>\widehat{DE}+2\cdot \widehat{HJ} + 3\cdot\widehat{FG},</math> where the arcs are measured in degrees. |
<asy> | <asy> | ||
import olympiad; | import olympiad; | ||
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</asy> | </asy> | ||
− | == Solution == | + | == Solution 1== |
Notice that due to midpoints, <math>\triangle DEF\sim\triangle FBD\sim\triangle AFE\sim\triangle EDC\sim\triangle ABC</math>. As a result, the angles and arcs are readily available. Due to inscribed angles, | Notice that due to midpoints, <math>\triangle DEF\sim\triangle FBD\sim\triangle AFE\sim\triangle EDC\sim\triangle ABC</math>. As a result, the angles and arcs are readily available. Due to inscribed angles, | ||
<cmath>\widehat{DE}=2\angle DFE=2\angle ACB=2\cdot36=72^\circ</cmath> | <cmath>\widehat{DE}=2\angle DFE=2\angle ACB=2\cdot36=72^\circ</cmath> | ||
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~ Edited by [https://artofproblemsolving.com/wiki/index.php/User:Aoum aoum] | ~ Edited by [https://artofproblemsolving.com/wiki/index.php/User:Aoum 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*} | ||
+ | |||
+ | ~ [https://artofproblemsolving.com/community/user/1096232 Pengu14] | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | [[File:2025AIMEIIP5.png|415px|right]] | ||
+ | |||
+ | 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]] | ||
==See also== | ==See also== |
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.