HMMT November, 2024

Geometry Difficulty 4.6 AIME Find the answer United States

Problem:

Let $ABCD$ be a rectangle whose vertices are labeled in counterclockwise order with $AB=32$ and $AD=60$ . Rectangle $AB^{\prime}C^{\prime}D^{\prime}$ is constructed by rotating $ABCD$ counterclockwise about $A$ by $60^{\circ}$ . Given that lines $BB^{\prime}$ and $DD^{\prime}$ intersect at point $X$ , compute $CX$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:

The key claim is the following.
Claim 1. $\angle BXD = 90^{\circ}$ .

Proof. We see that $\angle ABB^{\prime} = \angle AD^{\prime}D = 60^{\circ}$ and $\angle BAD^{\prime} = 90^{\circ} + \angle DAD^{\prime} = 150^{\circ}$ , from which we get $\angle BX D^{\prime} = \angle BXD = 90^{\circ}$ . Note that the fact that $BB^{\prime}$ and $DD^{\prime}$ are perpendicular is true regardless of how much we rotate the rectangle.

Now since $\angle BAD = \angle BXD$ , we establish that $X$ lies on the circumcircle of $ABCD$ , which has diameter $\sqrt{32^{2} + 60^{2}} = 68$ . Moreover, we have $\angle CDX = 90^{\circ} + \angle ADD^{\prime} = 150^{\circ}$ , so we discover that $CX = 34$ by applying the extended law of sines to $\triangle CDX$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

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