The 80-80-20 Triangle Problem, A Derivative, Solution #1

Manifesto | Index | What's new | rss feed

| Bookmark |

| Follow us

| Recommend | Contact |

ABC is an isosceles triangle with vertex angle ∠BAC = 20° and AB = AC. Point E is on AB such that AE = BC. Find the measure of ∠AEC.

Solution

The solution is by Vladimir Dubrovsky and, besides [Honsberger], it appeared in the Russian Kvant and subsequently in the now defunct Quantum magazine (May-June 1994).

Imagine a circle with center A and radius AB = AC. B is away from C by an arc subtending an angle of 20°. Step twice in this direction by the same angular measure getting points U and V on the circle. Naturally,

AC = AB = AU = AV

and

∠BAC = ∠BAU = ∠UAV = 20°

But this means that ∠CAV = 60° and AC = AV making ΔCAV equilateral, so that CV = AC.

Inscribed ∠CVU is subtended by an arc of 40° and is, therefore, equal 20°:

∠CVU = 20°.

In addition, UV = BU = BC. Consider now triangles AEC and VUC. They are equal by SAS: AE = UV, AC = CV, and ∠CVU = 20° = ∠CAE. But in ΔVUC, ∠VCU = 10° (as an inscribed angle subtended by a 20° arc). It follows that ∠ACE = 10° and ∠AEC = 180° - 10° - 20° = 150°.

Reference

R. Honsberger, Mathematical Chestnuts from Around the World, MAA, 2001, Ch 2.

40617822