In a triangle PQR, P is the largest angle and cos P = 1/3. Further the incircle of the triangle touc

Q: Wait, why do we assume $P$ is the largest angle? Do I need to prove it?

Great question! You don't need to prove it, but you must verify it. The problem states $P$ is the largest angle, which implies that the side opposite to it, $QR$, must be the longest side. By expressing all sides in terms of $x$, we found $QR = 2x+6$, which is indeed greater than $PQ = 2x+2$ and $PR = 2x+4$. This confirms our setup is consistent with the problem statement.

Q: Why can't we just use the Sine Rule instead of the Cosine Rule?

The Sine Rule relates sides to the sines of their opposite angles (e.g., $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$). While valid, we don't know the values of angles $Q$ or $R$, nor their sines. The Cosine Rule is far more powerful here because it connects all three sides directly to the cosine of the known angle $P$, allowing us to solve for $x$ without needing the other angles.

Q: Is there a constraint on $x$ being an even integer?

Yes, absolutely! The problem explicitly states that the lengths $PN, QL,$ and $RM$ are consecutive even integers. This means $x$ must be an even integer. When we solved the equation and found $x=8$, we checked this condition. Since 8 is an even integer, our solution is valid. Always keep an eye on such constraints; they are often the key to rejecting extraneous roots in more complex problems.

Question

In a triangle $PQR, P$ is the largest angle and $\cos P = 1/3$. Further the incircle of the triangle touches the sides $PQ, QR$ and $RP$ at $N, L$ and $M$ respectively, such that the lengths of $PN, QL$ and $RM$ are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are)

Options:

(A) 16

(B) 18

(C) 24

(D) 22

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome to this beautiful exploration of geometry. Today, we are not just solving a problem; we are peeling back the layers of a triangle to reveal the elegant relationships hidden within its incircle.

Imagine you are standing at the vertices of triangle $PQR$. You see an incircle, a circle perfectly nestled inside, touching the sides at points $N, L,$ and $M$.

The first step in our journey is to invoke the **Equal Tangents Theorem**. From any external point, the two tangents drawn to a circle are equal in length.

Let us define the tangent from vertex $P$ as $x$. Thus, $PN = PM = x$. Similarly, let the tangent from $Q$ be $y$, so $QN = QL = y$. And from $R$, let the tangent be $z$, so $RL = RM = z$.

### The Rhythm of Consecutive Integers

The problem gives us a rhythmic clue: $PN, QL,$ and $RM$ are consecutive even integers. We have already set $PN = x$. Therefore, $QL = x+2$ and $RM = x+4$.

Because $x$ is the length of a tangent, it must be an even integer. Now, look at the sides of the triangle:

$$PQ = PN + QN = x + (x+2) = 2x+2$$

$$QR = QL + LR = (x+2) + (x+4) = 2x+6$$

$$PR = PM + MR = x + (x+4) = 2x+4$$

We have successfully reduced the entire geometry of the triangle to a single variable, $x$.

### The Power of the Cosine Rule

Now, we must bridge the gap between these side lengths and the angle $P$. We are given $\cos P = \frac{1}{3}$.

The **Cosine Rule** is our bridge:

$$\cos P = \frac{b^2 + c^2 - a^2}{2bc}$$

Here, $a$ is the side opposite to $P$, which is $QR = 2x+6$. The other two sides are $b = PR = 2x+4$ and $c = PQ = 2x+2$.

Notice that $QR$ is the longest side, which perfectly aligns with the problem's statement that $P$ is the largest angle. This is the beauty of consistency in mathematics.

### The Algebraic Dance

Now, let us substitute these expressions into our formula:

$$\frac{(2x+4)^2 + (2x+2)^2 - (2x+6)^2}{2(2x+4)(2x+2)} = \frac{1}{3}$$

This looks intimidating, but do not panic. Let us factor out the constants. Notice that every term in the numerator has a factor of $2^2 = 4$, and the denominator has a factor of $2 	imes 2 = 4$. They cancel out beautifully!

We are left with:

$$\frac{(x+2)^2 + (x+1)^2 - (x+3)^2}{2(x+2)(x+1)} = \frac{1}{3}$$

Expanding the numerator, we get $(x^2 + 4x + 4) + (x^2 + 2x + 1) - (x^2 + 6x + 9)$. Combining these terms, the $x^2$ terms simplify to $x^2$, the $x$ terms cancel out completely ($4x + 2x - 6x = 0$), and the constants become $4 + 1 - 9 = -4$.

The numerator is simply $x^2 - 4$.

### The Final Revelation

We are left with the following equation:

$$\frac{x^2 - 4}{2(x+2)(x+1)} = \frac{1}{3}$$

Recognizing that $x^2 - 4$ is a difference of squares, we factor it as $(x-2)(x+2)$. The $(x+2)$ terms cancel out, leaving us with:

$$\frac{x-2}{2(x+1)} = \frac{1}{3}$$

Cross-multiplying gives $3(x-2) = 2(x+1)$, which simplifies to $3x - 6 = 2x + 2$. Solving for $x$, we find **$x = 8$**.

Substituting this back, we find the side lengths: $PQ = 2(8)+2 = 18$, $QR = 2(8)+6 = 22$, and $PR = 2(8)+4 = 20$. You have conquered the problem!

Select Answer:

Visualized Solution

Visualizing the Incircle and Tangents

Defining Consecutive Even Integers

Expressing Side Lengths in terms of $x$

Identifying the Largest Angle

Applying the Cosine Rule

Raw Setup (Substitution)

Factoring out Constants

Expanding the Squares

Simplifying the Numerator

Factoring and Canceling

Solving for $x$

Calculating the Final Side Lengths

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Rhythm of Consecutive Integers

The Power of the Cosine Rule

The Algebraic Dance

The Final Revelation

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