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Difference between revisions of "2025 AIME II Problems/Problem 5"

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== Solution ==
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== 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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~ [https://artofproblemsolving.com/community/user/1096232 Pengu14]
 
~ [https://artofproblemsolving.com/community/user/1096232 Pengu14]
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==Solution 2==
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[[File:2025AIMEIIP5.png|400px|right]]
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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>
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~[[User:grogg007|grogg007]]
  
 
==See also==
 
==See also==

Revision as of 11:26, 26 July 2025

Problem

Suppose $\triangle ABC$ has angles $\angle BAC = 84^\circ, \angle ABC=60^\circ,$ and $\angle ACB = 36^\circ.$ Let $D, E,$ and $F$ be the midpoints of sides $\overline{BC}, \overline{AC},$ and $\overline{AB},$ respectively. The circumcircle of $\triangle DEF$ intersects $\overline{BD}, \overline{AE},$ and $\overline{AF}$ at points $G, H,$ and $J,$ respectively. The points $G, D, E, H, J,$ and $F$ divide the circumcircle of $\triangle DEF$ into six minor arcs, as shown. Find $\widehat{DE}+2\cdot \widehat{HJ} + 3\cdot\widehat{FG},$ 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]

Solution 1

Notice that due to midpoints, $\triangle DEF\sim\triangle FBD\sim\triangle AFE\sim\triangle EDC\sim\triangle ABC$. As a result, the angles and arcs are readily available. Due to inscribed angles, \[\widehat{DE}=2\angle DFE=2\angle ACB=2\cdot36=72^\circ\] Similarly, \[\widehat{FG}=2\angle FDB=2\angle ACB=2\cdot36=72^\circ\]

In order to calculate $\widehat{HJ}$, we use the fact that $\angle BAC=\frac{1}{2}(\widehat{FDE}-\widehat{HJ})$. We know that $\angle BAC=84^\circ$, and \[\widehat{FDE}=360-\widehat{FE}=360-2\angle FDE=360-2\angle CAB=360-2\cdot84=192^\circ\]

Substituting,

\begin{align*} 84 &= \frac{1}{2}(192-\widehat{HJ}) \\ 168 &= 192-\widehat{HJ} \\ \widehat{HJ} &= 24^\circ \end{align*}

Thus, $\widehat{DE}+2\cdot\widehat{HJ}+3\cdot\widehat{FG}=72+48+216=\boxed{336}^\circ$.

~ eevee9406

~ 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

2025AIMEIIP5.png

Notice that $\triangle ABC \sim \triangle FHE \sim \triangle DEF$ because $FE \parallel BC$ and $DE \parallel AB,$ so all angles in each triangle will be equal. Therefore, we have $\widehat{DE} = 2 \cdot 36^\circ = 72^\circ.$ Now, quadrilateral $FHED$ is cyclic, so opposite angles add to $180^\circ.$ Since we know from similar triangles that $\angle{FED} = 60^\circ + 36^\circ = 96^\circ,$ we see that $\angle{HFD} = 84^\circ.$ We also know that $\angle{JFE} = 60^\circ + 36^\circ = 96^\circ,$ so that means $\angle{JFH} = 96^\circ - 84^\circ = 12^\circ \implies \widehat{JH} = 24^\circ.$ Finally, $\angle{B} = 60^\circ = \frac{\widehat{JED} - \widehat{FG}}{2}.$ $\widehat{JED} = \widehat{DE} + \widehat{JE} = 72^\circ + 2(\angle{JFE}) = 192^\circ.$ So $\widehat{FG} = 192 - 120 = 72^\circ.$ The answer is $72 + 2\cdot 24 + 3\cdot 72 = \boxed{336}.$

~grogg007

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. AMC Logo.png

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