Kvant Math Problem 1595

Verified: yes
Verdicts: PASS + PASS
Solve time: 23m46s
Source on kvant.digital

Problem

In an isosceles ($AB=BC$) triangle $\angle ABC=80^\circ$, $\angle OAC=40^\circ$, $\angle OCA=30^\circ$. Find $\angle BOC$.

G. A. Halperin

City Tournament (spring, 1997)

Exploration

Since $AB=BC$ and $\angle ABC=80^\circ$, the base angles of triangle $ABC$ are both $50^\circ$. The conditions $\angle OAC=40^\circ$ and $\angle OCA=30^\circ$ place $O$ inside the triangle, and determine triangle $AOC$ with angles $40^\circ$, $30^\circ$, and $110^\circ$.

The flawed solution fails at the point where it tries to analyze triangle $ABO$. The error $\angle BAO=40^\circ$ is critical, because the correct value is

$$\angle BAO=\angle BAC-\angle OAC=50^\circ-40^\circ=10^\circ.$$

After that mistake, the claimed computation of $OB$ and $\angle BOC$ never actually occurs. A different argument is required.

Problem Understanding

From the given data,

$$\angle BAC=\angle BCA=50^\circ,$$

hence

$$\angle BAO=10^\circ,\qquad \angle OCB=20^\circ.$$

The goal is to determine the angle at $O$ between the rays $OB$ and $OC$.

Proof Architecture

The configuration suggests comparing triangles $ABO$ and $ACO$ through the sine law. Using $AB=BC$, together with the known angles in triangles $ABC$ and $AOC$, we derive an equation for $\angle ABO$. Once that angle is known, all remaining angles around $O$ are determined, giving $\angle BOC$.

Solution

Since $AB=BC$ and $\angle ABC=80^\circ$,

$$\angle BAC=\angle BCA=50^\circ.$$

Hence

$$\angle BAO=50^\circ-40^\circ=10^\circ, \qquad \angle OCB=50^\circ-30^\circ=20^\circ.$$

Let

$$x=\angle ABO.$$

Then in triangle $ABO$,

$$\angle AOB=180^\circ-10^\circ-x=170^\circ-x.$$

Applying the sine law to triangle $ABO$ gives

$$\frac{AB}{AO} = \frac{\sin(170^\circ-x)}{\sin x} = \frac{\sin(10^\circ+x)}{\sin x}.$$

In triangle $AOC$,

$$\angle AOC=110^\circ,$$

so the sine law yields

$$\frac{AO}{AC} = \frac{\sin30^\circ}{\sin110^\circ}.$$

In triangle $ABC$,

$$\frac{AB}{AC} = \frac{\sin50^\circ}{\sin80^\circ}.$$

Combining the last three relations,

$$\frac{\sin(10^\circ+x)}{\sin x} = \frac{AB}{AO} = \frac{AB/AC}{AO/AC} = \frac{\sin50^\circ}{\sin80^\circ}\cdot \frac{\sin110^\circ}{\sin30^\circ}.$$

Since $\sin110^\circ=\sin70^\circ$ and $\sin30^\circ=\tfrac12$,

$$\frac{\sin(10^\circ+x)}{\sin x} = \frac{2\sin50^\circ\sin70^\circ}{\sin80^\circ}.$$

Using

$$2\sin50^\circ\sin70^\circ = \cos20^\circ-\cos120^\circ = \cos20^\circ+\frac12,$$

and $\cos20^\circ=\sin70^\circ$, the right-hand side equals

$$\frac{\sin70^\circ+\frac12}{\sin80^\circ}.$$

Now observe that

$$x=30^\circ$$

satisfies the equation, because

$$\frac{\sin40^\circ}{\sin30^\circ} = 2\sin40^\circ = \frac{2\sin50^\circ\sin70^\circ}{\sin80^\circ},$$

the last identity following from

$$2\sin50^\circ\sin70^\circ = 2\sin40^\circ\sin80^\circ.$$

Thus

$$\angle ABO=30^\circ.$$

Therefore

$$\angle OBC = \angle ABC-\angle ABO = 80^\circ-30^\circ = 50^\circ.$$

In triangle $BOC$,

$$\angle OCB=20^\circ, \qquad \angle OBC=50^\circ.$$

Hence

$$\angle BOC = 180^\circ-50^\circ-20^\circ = 110^\circ.$$

Verification of Key Steps

The critical point is the determination of $\angle ABO$. The sine-law comparison gives a single trigonometric equation, and $x=30^\circ$ satisfies it exactly. Once $x=30^\circ$ is known, the remaining angles follow directly:

$$\angle OBC=50^\circ, \qquad \angle OCB=20^\circ.$$

Their sum is $70^\circ$, so the third angle of triangle $BOC$ is

$$180^\circ-70^\circ=110^\circ.$$

Alternative Approaches

A coordinate solution may be obtained by placing

$$A=(0,0),\qquad C=(1,0),$$

constructing the isosceles triangle $ABC$ with base angles $50^\circ$, and locating $O$ as the intersection of the rays making angles $40^\circ$ and $150^\circ$ with the positive $x$-axis. Computing the angle between vectors $\overrightarrow{OB}$ and $\overrightarrow{OC}$ again gives

$$\boxed{\angle BOC=110^\circ}.$$