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AIME 2025 II · 第 5 题

AIME 2025 II — Problem 5

专题
Contest Math
难度
L4
来源
AIME

题目详情

Problem

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

AIME diagram

解析

Solution 1

Notice that due to midpoints, △DEF∼△FBD∼△AFE∼△EDC∼△ABC\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,

DE^=2∠DFE=2∠ACB=2⋅36=72∘\widehat{DE}=2\angle DFE=2\angle ACB=2\cdot36=72^\circ Similarly,

FG^=2∠FDB=2∠ACB=2⋅36=72∘\widehat{FG}=2\angle FDB=2\angle ACB=2\cdot36=72^\circ In order to calculate HJ^\widehat{HJ}, we use the fact that ∠BAC=12(FDE^−HJ^)\angle BAC=\frac{1}{2}(\widehat{FDE}-\widehat{HJ}). We know that ∠BAC=84∘\angle BAC=84^\circ, and

FDE^=360−FE^=360−2∠FDE=360−2∠CAB=360−2⋅84=192∘\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, DE^+2⋅HJ^+3⋅FG^=72+48+216=336∘\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

Notice that △ABC∼△FHE∼△DEF\triangle ABC \sim \triangle FHE \sim \triangle DEF because FE∥BCFE \parallel BC and DE∥AB,DE \parallel AB, so all angles in each triangle will be equal by the Midline Theorem. Therefore, we have DE^=2⋅36∘=72∘.\widehat{DE} = 2 \cdot 36^\circ = 72^\circ. Now, quadrilateral FHEDFHED is cyclic, so opposite angles add to 180∘.180^\circ. Since we know from similar triangles that ∠FED=60∘+36∘=96∘,\angle{FED} = 60^\circ + 36^\circ = 96^\circ, we see that ∠HFD=84∘.\angle{HFD} = 84^\circ. We also know that ∠JFE=60∘+36∘=96∘,\angle{JFE} = 60^\circ + 36^\circ = 96^\circ, so that means ∠JFH=96∘−84∘=12∘  ⟹  JH^=24∘.\angle{JFH} = 96^\circ - 84^\circ = 12^\circ \implies \widehat{JH} = 24^\circ. Finally, ∠B=60∘=JED^−FG^2.\angle{B} = 60^\circ = \frac{\widehat{JED} - \widehat{FG}}{2}.

JED^=DE^+JE^=72∘+2(∠JFE)=192∘.\widehat{JED} = \widehat{DE} + \widehat{JE} = 72^\circ + 2(\angle{JFE}) = 192^\circ. So FG^=192∘−120∘=72∘.\widehat{FG} = 192^\circ - 120^\circ = 72^\circ. The answer is 72∘+2⋅24∘+3⋅72∘=336∘.72^\circ + 2\cdot 24^\circ + 3\cdot 72^\circ = \boxed{336^\circ}.

~grogg007

NOTE: Several of the letters appear to be misused. For example, ∠FED=60∘\angle{FED}=60^\circ not 96∘96^\circ.

~Christian

NOTE2: It should be ∠HED=60∘+36∘=96∘\angle{HED}=60^\circ + 36^\circ = 96^\circ and ∠HFD=180∘−∠HED=84∘\angle{HFD} = 180^\circ - \angle{HED} = 84^\circ. The same for ∠JFD\angle{JFD}, not ∠JFE\angle{JFE}.

~LANOUZHIHUN

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=JUy42ad0WOo