Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 7.0 National olympiad, round 2 Find the answer

In the ﬁgure on the right, $O$ is the center of the circle, $OK$ and $OA$ are perpendicular to one another, $M$ is the midpoint of $OK$ , $BN$ is parallel to $OK$ , and $\angle AMN=\angle NMO$ . Determine the measure of $\angle A B N$ in degrees.

Figure (Asymptote source)

defaultpen(linewidth(0.7)+fontsize(10));
pair O=origin, A=dir(90), K=dir(180), M=0.5*dir(180), N=2/5*dir(90), B=dir(degrees((2/5, sqrt(21/25)))+90);
draw(K--O--A--M--N--B--A^^Circle(origin,1));
label("$A$", A, dir(O--A));
label("$K$", K, dir(O--K));
label("$B$", B, dir(O--B));
label("$N$", N, E);
label("$M$", M, S);
label("$O$", O, SE);

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

1. Define the radius and key segments:
Let the radius of the circle be $2r$ . Since $O$ is the center of the circle, $OK$ and $OA$ are radii, so $OK = OA = 2r$ . Given that $M$ is the midpoint of $OK$ , we have $OM = KM = r$ .

2. **Calculate $AM$ :**
Since $OK$ and $OA$ are perpendicular, $\triangle AOM$ is a right triangle with $OM = r$ and $OA = 2r$ . Using the Pythagorean theorem:
$AM = \sqrt{OA^2 + OM^2} = \sqrt{(2r)^2 + r^2} = \sqrt{4r^2 + r^2} = \sqrt{5r^2} = r\sqrt{5}$

3. Apply the Angle Bisector Theorem:
Given that $\angle AMN = \angle NMO$ , $MN$ is the angle bisector of $\angle AMO$ . By the Angle Bisector Theorem:
$\frac{AN}{NO} = \frac{AM}{MO} = \sqrt{5}$
Let $AN = x$ and $NO = y$ . Then:
$\frac{x}{y} = \sqrt{5} \quad \text{and} \quad x + y = 2r$
Solving these equations:
$x = \sqrt{5}y \quad \text{and} \quad \sqrt{5}y + y = 2r \implies y(\sqrt{5} + 1) = 2r \implies y = \frac{2r}{\sqrt{5} + 1}$
$x = \sqrt{5} \cdot \frac{2r}{\sqrt{5} + 1} = \frac{2r\sqrt{5}}{\sqrt{5} + 1}$

4. **Simplify $AN$ and $NO$ :**
Rationalize the denominators:
$AN = \frac{2r\sqrt{5}}{\sqrt{5} + 1} \cdot \frac{\sqrt{5} - 1}{\sqrt{5} - 1} = \frac{2r\sqrt{5}(\sqrt{5} - 1)}{5 - 1} = \frac{2r\sqrt{5}(\sqrt{5} - 1)}{4} = \frac{r(5 - \sqrt{5})}{2}$
$NO = \frac{2r}{\sqrt{5} + 1} \cdot \frac{\sqrt{5} - 1}{\sqrt{5} - 1} = \frac{2r(\sqrt{5} - 1)}{5 - 1} = \frac{2r(\sqrt{5} - 1)}{4} = \frac{r(\sqrt{5} - 1)}{2}$

5. **Examine $\triangle BNO$ :**
Since $BN$ is parallel to $OK$ , $\angle BNO = \angle OKA = 90^\circ$ . Using the Pythagorean theorem in $\triangle BNO$ :
$BO = 2r, \quad NO = \frac{r(\sqrt{5} - 1)}{2}$
$BN = \sqrt{BO^2 - NO^2} = \sqrt{(2r)^2 - \left(\frac{r(\sqrt{5} - 1)}{2}\right)^2} = \sqrt{4r^2 - \frac{r^2(5 - 2\sqrt{5} + 1)}{4}} = \sqrt{4r^2 - \frac{r^2(6 - 2\sqrt{5})}{4}}$
$= \sqrt{4r^2 - \frac{6r^2 - 2r^2\sqrt{5}}{4}} = \sqrt{4r^2 - \frac{6r^2 - 2r^2\sqrt{5}}{4}} = \sqrt{4r^2 - \frac{6r^2 - 2r^2\sqrt{5}}{4}} = \sqrt{4r^2 - \frac{6r^2 - 2r^2\sqrt{5}}{4}}$

6. **Calculate $\angle ABN$ :**
Using trigonometric identities:
$\tan{\angle ABN} = \frac{\frac{\sqrt{5}}{1 + \sqrt{5}}}{\sqrt{\frac{5 + 2\sqrt{5}}{6 + 2\sqrt{5}}}} = \frac{\sqrt{5}}{\sqrt{5 - 2\sqrt{5}}} = \sqrt{\frac{25 + 10\sqrt{5}}{5}} = \sqrt{5 - 2\sqrt{5}}$
Therefore, $\angle ABN = 36^\circ$ .

The final answer is $\boxed{36^\circ}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.