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

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== Solution ==
 
== Solution ==
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Notice that due to midpoints, <math>\triangle DEF\cong\triangle FBD\cong\triangle AFE\cong\triangle EDC\cong\triangle ABC</math>. As a result, the angles and arcs are readily available. Due to inscribed angles,
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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>
 
Similarly,
 
Similarly,

Revision as of 03:18, 14 February 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

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, \[84=\frac{1}{2}(192-\widehat{HJ})\] \[168=192-\widehat{HJ}\] \[\widehat{HJ}=24^\circ\]

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

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