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, 2023

Geometry Difficulty 5.6 AIME, harder Prove it United States

Problem:

Elisenda has a piece of paper in the shape of a triangle with vertices $A$ , $B$ , and $C$ such that $AB = 42$ . She chooses a point $D$ on segment $AC$ , and she folds the paper along line $BD$ so that $A$ lands at a point $E$ on segment $BC$ . Then, she folds the paper along line $DE$ . When she does this, $B$ lands at the midpoint of segment $DC$ . Compute the perimeter of the original unfolded triangle.

Solution

Solution:

Let $F$ be the midpoint of segment $DC$ .

Evidently $\angle ADB = 60^{\circ} = \angle BDE = \angle EDC$ . Moreover, we have $BD = DF = FC$ , $AD = DE$ , and $AB = BE$ . Hence angle bisector on $BDC$ gives us that $BE = 42$ , $EC = 84$ , and hence angle bisector on $ABC$ gives us that if $AD = x$ then $CD = 3x$ . Now this gives $BD = 3x/2$ , so thus the Law of Cosines on $ADB$ gives $x = 12\sqrt{7}$ .

Hence, $BC = 42 + 84 = 126$ and $AC = 4x = 48\sqrt{7}$ . The answer is $42 + 126 + 48\sqrt{7} = 168 + 48\sqrt{7}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.