Let tan α, tan β and tan γ; α, β, γ ≠ (2n-1)π/2, n ∈ N be the slopes of three line segments OA OB an

Q: Wait, why is the orthocenter $\vec{H} = \vec{A} + \vec{B} + \vec{C}$? I thought the orthocenter was just the intersection of altitudes.

That is a fantastic question! You are absolutely right that the orthocenter is the intersection of altitudes. However, there is a beautiful, lesser-known theorem in geometry: if the circumcenter of a triangle is at the origin, the position vector of the orthocenter is indeed the vector sum of the position vectors of the vertices. This is a powerful tool in competitive geometry because it turns a complex geometric construction into simple vector addition!

Q: Why does the orthocenter lying on the y-axis mean the sum of cosines is zero?

Think about the coordinates of the orthocenter $H$. We know $\vec{H} = \vec{A} + \vec{B} + \vec{C}$. If $A = (R \cos \alpha, R \sin \alpha)$, $B = (R \cos \beta, R \sin \beta)$, and $C = (R \cos \gamma, R \sin \gamma)$, then the x-coordinate of $H$ is $R(\cos \alpha + \cos \beta + \cos \gamma)$. If $H$ lies on the y-axis, its x-coordinate must be zero. Since the radius $R$ is not zero, the sum of the cosines must be zero. It is all about translating geometric constraints into algebraic ones!

Q: Is there a risk that the denominator $\cos \alpha \cos \beta \cos \gamma$ becomes zero?

The problem statement explicitly gives the condition $\alpha, \beta, \gamma \neq \frac{(2n-1)\pi}{2}$. This ensures that $\cos \alpha, \cos \beta, \cos \gamma$ are all non-zero, so the denominator is safe. Always check the domain constraints in JEE problems; they are there for a reason!

Question

Let $	an \alpha, 	an \beta$ and $	an \gamma; \alpha, \beta, \gamma 
eq \frac{(2n-1)\pi}{2}, n \in N$ be the slopes of three line segments OA OB and OC, respectively, where O is origin. If circumcentre of $	riangle ABC$ coincides with origin and its orthocentre lies on y -axis, then the value of $\left(\frac{\cos 3\alpha+\cos 3\beta+\cos 3\gamma}{\cos \alpha \cos \beta \cos \gamma}ight)^2$ is equal to :

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing at the origin of a coordinate plane, looking out at a triangle $ABC$ inscribed in a circle centered at the origin. When the circumcenter of a triangle coincides with the origin, the geometry becomes incredibly elegant.

We represent the vertices $A, B,$ and $C$ using polar coordinates. If the circumradius is $R$, then the vertices are:
$A = (R \cos \alpha, R \sin \alpha)$
$B = (R \cos \beta, R \sin \beta)$
$C = (R \cos \gamma, R \sin \gamma)$

The slopes of the lines $OA, OB,$ and $OC$ are given as $	an \alpha, 	an \beta,$ and $	an \gamma$, which perfectly aligns with our polar representation.

### The Vector Secret of the Orthocenter

In any triangle where the circumcenter is at the origin, the orthocenter $H$ has a position vector $\vec{H}$ that is simply the sum of the position vectors of the vertices:
$$\vec{H} = \vec{A} + \vec{B} + \vec{C}$$

By summing the coordinates, we find that the x-coordinate of the orthocenter is $H_x = R(\cos \alpha + \cos \beta + \cos \gamma)$. The problem states that the orthocenter lies on the y-axis, which implies the x-coordinate must be zero.

Since $R 
eq 0$, we arrive at the crucial condition:
$$\cos \alpha + \cos \beta + \cos \gamma = 0$$

### The Algebraic Bridge

With $\cos \alpha + \cos \beta + \cos \gamma = 0$ established, we invoke a classic algebraic identity. If $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$.

Applying this to our cosine terms, we obtain:
$$\cos^3 \alpha + \cos^3 \beta + \cos^3 \gamma = 3 \cos \alpha \cos \beta \cos \gamma$$

This identity serves as the bridge that connects our geometric constraint to the trigonometric expression we need to evaluate.

### The Triple Angle Transformation

We evaluate the expression involving $\cos 3\alpha, \cos 3\beta,$ and $\cos 3\gamma$ using the triple angle formula: $\cos 3	heta = 4\cos^3 	heta - 3\cos 	heta$. Summing these for all three angles, we get:

$$\sum \cos 3\alpha = 4(\cos^3 \alpha + \cos^3 \beta + \cos^3 \gamma) - 3(\cos \alpha + \cos \beta + \cos \gamma)$$

Substituting our known values—the sum of the cubes is $3 \cos \alpha \cos \beta \cos \gamma$ and the sum of the cosines is $0$—the expression simplifies dramatically:

$$\sum \cos 3\alpha = 4(3 \cos \alpha \cos \beta \cos \gamma) - 3(0) = 12 \cos \alpha \cos \beta \cos \gamma$$

### The Grand Finale

Finally, we look at the ratio we need to square:
$$\frac{\cos 3\alpha + \cos 3\beta + \cos 3\gamma}{\cos \alpha \cos \beta \cos \gamma}$$

Substituting our simplified numerator, we get:
$$\frac{12 \cos \alpha \cos \beta \cos \gamma}{\cos \alpha \cos \beta \cos \gamma} = 12$$

Squaring this result, we obtain $12^2 = \mathbf{144}$. This is a beautiful, clean result that emerges from the interplay of geometry and algebra.

Enter Numerical Value:

Visualized Solution

$V i s u a l i z in g t h e T r ian g l e A B C$

$D e f inin g P o l a r C oor d ina t es$

$T h e O r t h oce n t er P r o p er t y$

$C o n s t r ain t : O r t h oce n t er o n y - a x i s$

$A ppl y in g A l g e b r ai c I d e n t i t y$

$T r i pl e A n g l e F or m u l a$

$S u mmin g t h e T r i pl e A n g l es$

$A t o mi c S u b s t i t u t i o n$

$C a l c u l a t in g t h e R a t i o$

$F ina l R es u l t$

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

Analyzing the Setup

The Vector Secret of the Orthocenter

The Algebraic Bridge

The Triple Angle Transformation

The Grand Finale

Similar Questions

Let $A (6, 8)$ , $B (10 cos α, - 10 sin α)$ and $C (- 10 sin α, 10 cos α)$ , be the vertices of a triangle. If $L (a, 9)$ and $G (h, k)$ be its orthocenter and centroid respectively, then $(5 a - 3 h + 6 k + 100 sin 2 α)$ is equal to

For $t \in (0, 2 π)$ , if $A B C$ is an equilateral triangle with vertices $A (sin t, cos t), B (cos t, sin t)$ and $C (a, b)$ such that its orthocentre lies on a circle with centre $(1, \frac{1}{3})$ , then $(a^{2} - b^{2})$ is equal to :

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

Two vertices of a triangle $A B C$ are $A (3, - 1)$ and $B (- 2, 3)$ , and its orthocentre is $P (1, 1)$ . If the coordinates of the point $C$ are $(α, β)$ and the centre of the circle circumscribing the triangle $P A B$ is $(h, k)$ , then the value of $(α + β) + 2 (h + k)$ equals

Let the circumcentre of a triangle with vertices $A (a, 3), B (b, 5)$ and $C (a, b), ab > 0$ be $P (1, 1)$ . If the line $A P$ intersects the line $B C$ at the point $Q (k_{1}, k_{2})$ , then $k_{1} + k_{2}$ is equal to :

Let $A B C$ be the triangle such that the equations of lines $A B$ and $A C$ be $3 y - x = 2$ and $x + y = 2$ , respectively, and the points $B$ and $C$ lie on $x$ -axis. If $P$ is the orthocentre of the triangle $A B C$ , then the area of the triangle $P B C$ is equal to

If the points $P$ and $Q$ are respectively the circumcenter and the orthocentre of a $Δ A B C$ , then $P A + P B + P C$ is equal to _______

If the orthocentre of the triangle formed by the lines $2 x + 3 y - 1 = 0$ , $x + 2 y - 1 = 0$ and $a x + b y - 1 = 0$ , is the centroid of another triangle, whose circumcentre and orthocentre respectively are $(3, 4)$ and $(- 6, - 8)$ , then the value of $∣ a - b ∣$ is_______

The equations of the sides $A B, B C$ and $C A$ of a triangle $A B C$ are $2 x + y = 0$ , $x + p y = 15 a$ and $x - y = 3$ respectively. If its orthocentre is $(2, a)$ , $- \frac{1}{2} < a < 2$ , then $p$ is equal to

Consider a triangle having vertices $A (- 2, 3), B (1, 9)$ and $C (3, 8)$ . If a line $L$ passing through the circum-centre of triangle $A B C$ , bisects line $B C$ , and intersects y-axis at point $(0, \frac{α}{2})$ , then the value of real number $α$ is

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Triangle ABC

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Constraint: Orthocenter on y−axis

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Summing the Triple Angles

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

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

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