olymmath

Putnam 1999 B1

  • Session B

Right triangle ABCABC has right angle at CC and ∠BAC=θ\angle BAC =\theta; the point DD is chosen on ABAB so that ∣AC∣=∣AD∣=1|AC|=|AD|=1; the point EE is chosen on BCBC so that ∠CDE=θ\angle CDE = \theta. The perpendicular to BCBC at EE meets ABAB at FF. Evaluate lim⁡θ→0∣EF∣\lim_{\theta\rightarrow 0} |EF|.

Curated text Kedlaya Putnam archive. Quoted with source.

Solutions

Human solution Solution 1 by Manjul Bhargava, Kiran Kedlaya and Lenny Ng

The terms of the source let olymmath link to this solution but not copy it. Read the solution (Kedlaya Putnam archive).

Locator putnam/1999/b1 · Tag BSW7VQRJY3 · Edit this page on GitHub · Keys: j next, k previous