Let A(6,8), B(10cos α, -10sin α) and C(-10sin α, 10cos α), be the vertices of a triangle. If L(a,9)

Q: Wait, why is the circumcenter at the origin? Is that always true?

It is not always true, but in this specific problem, notice that the distance of each vertex from the origin is constant: $\sqrt{x^2 + y^2} = 10$. Since all three vertices are equidistant from the origin, the origin must be the center of the circle passing through them—the circumcircle.

Q: Why use the Euler line? Can't I just find the orthocenter using slopes?

You could, but that would be a nightmare of algebra involving slopes of altitudes and intersection points. The Euler line is a powerful geometric shortcut that links the orthocenter, centroid, and circumcenter, turning a complex coordinate geometry problem into a simple ratio problem.

Q: How did we get $\sin 2\alpha$ from $\cos\alpha - \sin\alpha$?

By squaring both sides! When you square $(\cos\alpha - \sin\alpha)$, you get $\cos^2\alpha + \sin^2\alpha - 2\sin\alpha\cos\alpha$. Since $\cos^2\alpha + \sin^2\alpha = 1$ and $2\sin\alpha\cos\alpha = \sin 2\alpha$, the expression simplifies to $1 - \sin 2\alpha$.

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of the Triangle

Imagine you are standing on a coordinate plane, looking at three points: $A(6,8)$, $B(10\cos\alpha, -10\sin\alpha)$, and $C(-10\sin\alpha, 10\cos\alpha)$.

First, we calculate the distance of vertex $A$ from the origin:
$$x^2 + y^2 = 6^2 + 8^2 = 36 + 64 = 100$$
The distance is $10$.

Now, consider vertices $B$ and $C$. Squaring their coordinates yields:
$$100(\cos^2\alpha + \sin^2\alpha) = 100$$
All three vertices lie on a circle of radius $10$ centered at the origin. This is our first breakthrough: the circumcenter $S$ is $(0,0)$.

### The Euler Line: A Hidden Connection

We need to find the orthocenter $L(a,9)$ and the centroid $G(h,k)$. In any triangle, the orthocenter $L$, the centroid $G$, and the circumcenter $S$ are collinear.

The centroid $G$ divides the segment $LS$ in a $2:1$ ratio. Using the section formula, we express $G$ as:
$$G = \frac{1 \cdot L + 2 \cdot S}{3}$$

Substituting our known points, we get:
$$G(h,k) = \frac{1(a,9) + 2(0,0)}{3} = \left(\frac{a}{3}, 3\right)$$

### The Algebraic Bridge

From this result, we immediately see that $k = 3$ and $h = \frac{a}{3}$, which implies $a = 3h$. We have successfully linked the orthocenter and the centroid.

Next, we use the definition of the centroid as the average of the vertices' coordinates. The $y$-coordinate of the centroid is:
$$k = \frac{y_A + y_B + y_C}{3}$$

Substituting our values:
$$3 = \frac{8 - 10\sin\alpha + 10\cos\alpha}{3}$$
Multiplying by $3$ gives $9 = 8 + 10(\cos\alpha - \sin\alpha)$, which simplifies to:
$$\cos\alpha - \sin\alpha = 0.1$$

### The Final Act

To find $\sin 2\alpha$, we square our previous result:
$$(\cos\alpha - \sin\alpha)^2 = (0.1)^2$$
$$1 - \sin 2\alpha = 0.01 \Rightarrow \sin 2\alpha = 0.99$$

Now, we determine $h$ using the $x$-coordinate of the centroid:
$$h = \frac{x_A + x_B + x_C}{3} = \frac{6 + 10\cos\alpha - 10\sin\alpha}{3}$$
Since we know $\cos\alpha - \sin\alpha = 0.1$, this becomes:
$$h = \frac{6 + 10(0.1)}{3} = \frac{7}{3}$$

Consequently, $a = 3h = 7$. Finally, we evaluate the expression:
$$5a - 3h + 6k + 100\sin 2\alpha = 5(7) - 3\left(\frac{7}{3}\right) + 6(3) + 100(0.99)$$
$$= 35 - 7 + 18 + 99 = 145$$

The journey is complete, and the final answer is **145**.

Animated Solution for Mathematics - Straight Lines: 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

Enter Numerical Value:

Visualized Solution

Analyzing Vertex $A (6, 8)$

Analyzing Vertices $B$ and $C$

Identifying the Circumcenter $S$

The Euler Line Property

Applying Section Formula for $G$

Finding $k$ and $h$ in terms of $a$

Using the Centroid Formula for $y$

Solving for $(cos α - sin α)$

Finding $sin 2 α$

Finding $h$ from the $x$ -coordinate

Calculating $h$ and $a$

Final Calculation

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

Analyzing the Geometry of the Triangle

The Euler Line

The Algebraic Bridge

The Final Act

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Enter Numerical Value:

Visualized Solution

Analyzing Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A(6,8)

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

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

The Euler Line Property

Applying Section Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G

Using the Centroid Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (cosα−sinα)

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin2α

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h from the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-coordinate

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