Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 7.1 National olympiad, round 2 Find the answer

Point $G$ is where the medians of the triangle $ABC$ intersect and point $D$ is the midpoint of side $BC$ . The triangle $BDG$ is equilateral with side length 1. Determine the lengths, $AB$ , $BC$ , and $CA$ , of the sides of triangle $ABC$ .

Figure (Asymptote source)

size(200);
defaultpen(fontsize(10));
real r=100.8933946;
pair A=sqrt(7)*dir(r), B=origin, C=(2,0), D=midpoint(B--C), E=midpoint(A--C), F=midpoint(A--B), G=centroid(A,B,C);
draw(A--B--C--A--D^^B--E^^C--F);
pair point=G;
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));
label("$E$", E, dir(point--E));
label("$F$", F, dir(point--F));
label("$G$", G, dir(20));
label("1", B--G, dir(150));
label("1", D--G, dir(30));
label("1", B--D, dir(270));

Your answer

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

Solution

1. Identify the given information and the properties of the centroid:
- Point $G$ is the centroid of triangle $ABC$ .
- Point $D$ is the midpoint of side $BC$ .
- Triangle $BDG$ is equilateral with side length 1.

2. **Determine the length of $BC$ :**
- Since $D$ is the midpoint of $BC$ and $BD = 1$ , it follows that $DC = 1$ .
- Therefore, $BC = BD + DC = 1 + 1 = 2$ .

3. Use the centroid properties:
- The centroid $G$ divides each median into a ratio of 2:1.
- Let $E$ be the midpoint of $AC$ , and $F$ be the midpoint of $AB$ .

4. **Calculate $BG$ and $GD$ :**
- Since $BDG$ is equilateral, $BG = GD = 1$ .

5. **Determine the length of $AG$ :**
- The centroid divides the median $AD$ in the ratio 2:1.
- Therefore, $AG = 2 \times GD = 2 \times 1 = 2$ .

6. **Calculate $GE$ :**
- Since $G$ is the centroid, $GE = \frac{1}{2} \times AG = \frac{1}{2} \times 2 = 1$ .

7. **Apply the Law of Cosines in $\triangle AGE$ :**
- $\angle AGE = 60^\circ$ because $BDG$ is equilateral.
- Using the Law of Cosines:
$AE^2 = AG^2 + GE^2 - 2 \cdot AG \cdot GE \cdot \cos(60^\circ)$
$AE^2 = 2^2 + 1^2 - 2 \cdot 2 \cdot 1 \cdot \frac{1}{2}$
$AE^2 = 4 + 1 - 2 = 3$
$AE = \sqrt{3}$

8. **Determine the length of $CA$ :**
- Since $E$ is the midpoint of $AC$ , $AE = \frac{1}{2} \times AC$ .
- Therefore, $AC = 2 \times AE = 2 \times \sqrt{3} = 2\sqrt{3}$ .

9. **Apply the Law of Cosines in $\triangle AGB$ :**
- $\angle AGB = 120^\circ$ because $BDG$ is equilateral.
- Using the Law of Cosines:
$AB^2 = AG^2 + BG^2 - 2 \cdot AG \cdot BG \cdot \cos(120^\circ)$
$AB^2 = 2^2 + 1^2 - 2 \cdot 2 \cdot 1 \cdot (-\frac{1}{2})$
$AB^2 = 4 + 1 + 2 = 7$
$AB = \sqrt{7}$

The final answer is $\boxed{ AB = \sqrt{7} }$ , $BC = 2$ , and $CA = 2\sqrt{3}$ .

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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.