Let the position vectors of three vertices of a triangle be 4p + q - 3r, -5p + q + 2r and 2p - q + 2

Q: Wait, why do we use the Euler Line? Can't we just find the perpendicular bisectors to get the circumcenter?

You absolutely could, but that would be a long, tedious path involving finding the equations of lines and their intersections. In JEE Advanced, time is your most precious resource. The Euler Line property is a powerful shortcut that turns a complex coordinate geometry problem into a simple vector algebra exercise. Always look for the symmetry first!

Q: Does the order of the ratio in the Euler Line formula matter? Is it $2:1$ or $1:2$?

That is a classic trap! The centroid $\vec{G}$ is always closer to the circumcenter $\vec{C}$ than to the orthocenter $\vec{H}$. The distance $HG$ is twice the distance $GC$. So, the ratio $HG:GC$ is $2:1$. When applying the section formula $\vec{G} = \frac{m\vec{C} + n\vec{H}}{m+n}$, ensure you map the ratios correctly: $\vec{G} = \frac{1\cdot\vec{H} + 2\cdot\vec{C}}{3}$.

Q: Is the Euler Line property valid for all triangles?

Yes, it is a universal property for all triangles! Whether it is acute, obtuse, or right-angled, the orthocenter, centroid, and circumcenter are always collinear, forming the Euler Line. The only exception is an equilateral triangle, where all these centers coincide at the same point, making the line undefined.

Question

Let the position vectors of three vertices of a triangle be $4\vec{p}+\vec{q}-3\vec{r}$, $-5\vec{p}+\vec{q}+2\vec{r}$ and $2\vec{p}-\vec{q}+2\vec{r}$ If the position vectors of the orthocenter and the circumcenter of the triangle are $\frac{\vec{p}+\vec{q}+\vec{r}}{4}$ and $\alpha\vec{p}+\beta\vec{q}+\gamma\vec{r}$ respectively, then $\alpha+2\beta+5\gamma$ is equal to:

Options:

(A) 3

(B) 4

(C) 1

(D) 6

Sigma Percentile · Accepted Answer

### Analyzing the Triangle's DNA

We begin with the given position vectors of the vertices:
$\vec{A} = 4\vec{p} + \vec{q} - 3\vec{r}$
$\vec{B} = -5\vec{p} + \vec{q} + 2\vec{r}$
$\vec{C}_{tri} = 2\vec{p} - \vec{q} + 2\vec{r}$

The centroid $\vec{G}$ is the average of these vertices:
$$\vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}_{tri}}{3}$$

Summing the coefficients of $\vec{p}$, $\vec{q}$, and $\vec{r}$ respectively:
$\sum \vec{V} = (4 - 5 + 2)\vec{p} + (1 + 1 - 1)\vec{q} + (-3 + 2 + 2)\vec{r} = \vec{p} + \vec{q} + \vec{r}$

Thus, the centroid is:
$$\vec{G} = \frac{\vec{p} + \vec{q} + \vec{r}}{3}$$

### The Euler Line Property

In any triangle, the orthocenter $\vec{H}$, the centroid $\vec{G}$, and the circumcenter $\vec{C}$ are collinear, forming the Euler line. The centroid $\vec{G}$ divides the segment $\vec{HC}$ internally in the ratio $2:1$.

Using the section formula, we express $\vec{G}$ as:
$$\vec{G} = \frac{1 \cdot \vec{H} + 2 \cdot \vec{C}}{3}$$

### The Algebraic Dance

Rearranging the section formula gives $3\vec{G} = \vec{H} + 2\vec{C}$. Given the orthocenter $\vec{H} = \frac{\vec{p} + \vec{q} + \vec{r}}{4}$, we substitute the known values:

$$3\left(\frac{\vec{p} + \vec{q} + \vec{r}}{3}\right) = \frac{\vec{p} + \vec{q} + \vec{r}}{4} + 2\vec{C}$$

Simplifying the left side, we obtain:
$$\vec{p} + \vec{q} + \vec{r} = \frac{\vec{p} + \vec{q} + \vec{r}}{4} + 2\vec{C}$$

Isolating $2\vec{C}$:
$$2\vec{C} = \left(1 - \frac{1}{4}\right)(\vec{p} + \vec{q} + \vec{r}) = \frac{3}{4}(\vec{p} + \vec{q} + \vec{r})$$

Dividing by two, the circumcenter is:
$$\vec{C} = \frac{3}{8}\vec{p} + \frac{3}{8}\vec{q} + \frac{3}{8}\vec{r}$$

### Final Calculation

Comparing this to the form $\vec{C} = \alpha\vec{p} + \beta\vec{q} + \gamma\vec{r}$, we identify:
$\alpha = \frac{3}{8}$, $\beta = \frac{3}{8}$, and $\gamma = \frac{3}{8}$.

We calculate the required value $\alpha + 2\beta + 5\gamma$:
$$\frac{3}{8} + 2\left(\frac{3}{8}\right) + 5\left(\frac{3}{8}\right) = \frac{3 + 6 + 15}{8} = \frac{24}{8} = 3$$

The final result is **3**.

Select Answer:

Visualized Solution

Visualize the Triangle

Recall Centroid Formula

Summing the Vertices

Finding Centroid $G$

The Euler Line Property

Setting up the Equation

Isolating the Circumcenter

Calculating $C$

Identifying Coefficients

Final Expression Setup

Final Calculation

Summary & Takeaway

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

Analyzing the Triangle's DNA

The Euler Line Property

The Algebraic Dance

Final Calculation

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Select Answer:

Visualized Solution

Visualize the Triangle

Recall Centroid Formula

Summing the Vertices

Finding Centroid .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G

The Euler Line Property

Setting up the Equation

Isolating the Circumcenter

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

Identifying Coefficients

Final Expression Setup

Final Calculation

Summary & Takeaway

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

Analyzing the Triangle's DNA

The Euler Line Property

The Algebraic Dance

Final Calculation

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Finding Centroid $G$

Calculating $C$

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