A triangle ABC has sides AB = AC = 5 cm and BC = 6 cm. Triangle A'B'C' is the reflection of the tria

Q: Wait, why is the distance $AA'$ exactly $4$ cm? I thought it was just $2$ cm because the line is $2$ cm away.

That is a classic trap! Remember the fundamental property of reflection: the image is as far behind the mirror as the object is in front. If the object $A$ is $2$ cm in front of the mirror $L_1$, the image $A'$ must be $2$ cm behind the mirror. The total distance $AA'$ is the sum of these two segments: $2 + 2 = 4$ cm. Always visualize the mirror as the midpoint of the segment connecting the object and the image.

Q: How did you get $12$ cm for $A'A''$? I keep getting $6$ cm.

You are likely forgetting the altitude of the triangle! The distance from $A'$ to the mirror $L_2$ is not just the $2$ cm gap. It is the distance from $A'$ to the base $B'C'$ (which is the altitude, $4$ cm) plus the distance from the base to the mirror ($2$ cm). So, the total distance from $A'$ to the mirror is $4 + 2 = 6$ cm. Since $A''$ is the reflection, it is $6$ cm on the other side. Thus, $A'A'' = 6 + 6 = 12$ cm.

Q: Why is the angle between the normals $180^\circ - \beta$? This feels like a leap.

It is a beautiful geometric result. The vector $\vec{AA'}$ is the normal to $AB$. The vector $\vec{A'A''}$ is the normal to $B'C'$. When you reflect a shape, you are essentially rotating and flipping it. The angle between the normals of two lines is equal to the supplement of the angle between the lines themselves. Since the base angle of our triangle is $\beta$, the angle between the normals is indeed $180^\circ - \beta$. Trust the geometry; it never lies!

Question

A triangle $ABC$ has sides $AB = AC = 5$ cm and $BC = 6$ cm. Triangle $A'B'C'$ is the reflection of the triangle $ABC$ in a line parallel to $AB$ placed at a distance $2$ cm from $AB$, outside the triangle $ABC$. Triangle $A''B''C''$ is the reflection of the triangle $A'B'C'$ in a line parallel to $B'C'$ placed at a distance $2$ cm from $B'C'$ outside the triangle $A'B'C'$. Find the distance between $A$ and $A''$.

Sigma Percentile · Accepted Answer

### Analyzing the Foundation

Imagine you are standing before a blank canvas. We start with an isosceles triangle, $\Delta ABC$, where $AB = AC = 5$ cm and the base $BC = 6$ cm.

Before we do anything else, we must understand the soul of this triangle. If we drop an altitude from vertex $A$ to the base $BC$, we bisect the base. This creates two right-angled triangles with a base of $3$ cm and a hypotenuse of $5$ cm.

Using the Pythagorean theorem, the altitude $h$ is:

$$h = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = 4 	ext{ cm}$$

Keep this number, $4$, close to your heart; it is the key to everything that follows.

### The First Mirror

Now, we introduce our first mirror, $L_1$. It is parallel to $AB$ and sits $2$ cm outside the triangle.

When we reflect point $A$ across this mirror, we create $A'$. Remember the golden rule of reflection: the mirror is the perpendicular bisector of the segment connecting the object and its image.

Since $A$ is $2$ cm from the mirror, $A'$ must be $2$ cm on the other side. Therefore, the total distance is:

$$AA' = 2 + 2 = 4 	ext{ cm}$$

The vector $\vec{AA'}$ is perpendicular to $AB$. We have successfully moved our first piece.

### The Second Mirror and the Trap

Here is where many students stumble. We reflect the entire triangle $\Delta ABC$ to get $\Delta A'B'C'$. Now, we place a second mirror, $L_2$, parallel to $B'C'$, at a distance of $2$ cm outside the triangle.

We need to find the reflection of $A'$ across this new mirror to get $A''$. Think carefully: what is the distance of $A'$ from $L_2$?

The altitude of our triangle is $4$ cm. The mirror is $2$ cm away from the base $B'C'$. Thus, the total distance from $A'$ to the mirror $L_2$ is $4 + 2 = 6$ cm.

Just like before, the reflection $A''$ will be $6$ cm on the other side of the mirror. This makes the total distance:

$$A'A'' = 6 + 6 = 12 	ext{ cm}$$

The vector $\vec{A'A''}$ is perpendicular to $B'C'$.

### The Vector Dance

We now have two vectors, $\vec{AA'}$ and $\vec{A'A''}$, with lengths $4$ cm and $12$ cm, respectively. To find the distance $AA''$, we need the angle between these two vectors.

Let $\beta$ be the base angle of our original triangle $\Delta ABC$. Since $\vec{AA'}$ is the normal to $AB$ and $\vec{A'A''}$ is the normal to $B'C'$, the angle between these two normals is $	heta = 180^\circ - \beta$.

Why? Because the angle between the normals of two lines is the supplement of the angle between the lines themselves. In our triangle, $\cos \beta = \frac{3}{5}$. Therefore, the cosine of our angle $	heta$ is:

$$\cos(180^\circ - \beta) = -\cos \beta = -\frac{3}{5}$$

### The Grand Finale

We have arrived at the final act. We have a triangle formed by points $A$, $A'$, and $A''$. We know two sides ($4$ and $12$) and the included angle ($	heta$). The Law of Cosines is our best friend here:

$$AA''^2 = AA'^2 + A'A''^2 - 2(AA')(A'A'') \cos(	heta)$$

Substituting our values:

$$AA''^2 = 4^2 + 12^2 - 2(4)(12)\left(-\frac{3}{5}ight)$$

Let us compute this with precision:

$$AA''^2 = 16 + 144 - 96\left(-\frac{3}{5}ight)$$
$$AA''^2 = 160 + \frac{288}{5}$$
$$AA''^2 = 160 + 57.6 = 217.6$$

Finally, we take the square root:

$$AA'' = \sqrt{217.6} = \sqrt{\frac{1088}{5}} = 8\sqrt{\frac{17}{5}} 	ext{ cm}$$

And there it is. The distance between $A$ and $A''$ is **$8\sqrt{\frac{17}{5}}$ cm**. You didn't just calculate a number; you tracked the movement of a point through space using the elegance of geometry.

Visualized Solution

Initial Setup: $Δ A B C$

Altitude of $Δ A B C$

First Mirror Line $L_{1}$

Distance $A A^{'}$

Reflected Triangle $A^{'} B^{'} C^{'}$

Altitude of $Δ A^{'} B^{'} C^{'}$

Second Mirror Line $L_{2}$

Distance $A^{'} A^{''}$

Angle Between Normals

Cosine of $∠ A A^{'} A^{''}$

Law of Cosines Setup

Substituting Values

Computing the Terms

Final Distance $A A^{''}$

The Sigma Insight: Properties of Triangles

Analyzing the Foundation

The First Mirror

The Second Mirror and the Trap

The Vector Dance

The Grand Finale

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Visualized Solution

Initial Setup: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔABC

Altitude of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔABC

First Mirror Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AA′

Reflected Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A′B′C′

Altitude of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔA′B′C′

Second Mirror Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A′A′′