Let ABC and PQR be any two triangles in the same plane. Assume that the perpendiculars from the poin

Q: Why do we sum the equations? Is that mathematically sound?

That is a brilliant question! We sum them because we are looking for a condition that holds true for the system as a whole. By summing, we create a linear combination that eliminates the specific point $H$, revealing an invariant property of the triangles themselves. It is a powerful technique in vector geometry to 'hide' a specific point to reveal the underlying relationship between the vertices.

Q: Wait, does this only work in 2D? The problem says 'in the same plane'.

Great observation! While the problem states they are in the same plane, the vector algebra actually holds in 3D space as well. The 'plane' constraint just makes it easier to visualize, but the beauty of vector algebra is its dimension-independence. You have just discovered a property that is more general than the problem statement implies!

Q: How does the final equation prove concurrency for the second set?

Recall that if $\vec{p} \cdot \vec{BC} + \vec{q} \cdot \vec{CA} + \vec{r} \cdot \vec{AB} = 0$, it implies the perpendiculars from $P, Q, R$ to the sides of $ABC$ satisfy the same dot product condition that defined the first concurrency. Since the condition is identical in form, it implies the existence of a point of concurrency for the second set as well. It is a beautiful, symmetric 'mirror' effect.

Question

Let $ABC$ and $PQR$ be any two triangles in the same plane. Assume that the perpendiculars from the points $A, B, C$ to the sides $QR, RP, PQ$ respectively are concurrent. Using vector methods or otherwise, prove that the perpendiculars from $P, Q, R$ to $BC, CA, AB$ respectively are also concurrent.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are uncovering a hidden symmetry in geometry.

We are looking at two triangles, $ABC$ and $PQR$, floating in a plane. We are told that the perpendiculars from $A$ to $QR$, $B$ to $RP$, and $C$ to $PQ$ are concurrent.

This is a special configuration, and we are going to prove that this property is reciprocal—that the perpendiculars from $P, Q, R$ to $BC, CA, AB$ are also concurrent. Let us dive in.

### Phase 1: The Vector Playground

First, let us ground ourselves. Geometry can be slippery, but vectors are our anchor.

Let the position vectors of the vertices of triangle $ABC$ be $\vec{a}, \vec{b}, \vec{c}$, and for triangle $PQR$, let them be $\vec{p}, \vec{q}, \vec{r}$.

By assigning these vectors, we have transformed a static geometric sketch into a dynamic algebraic system. We are no longer just drawing lines; we are manipulating the very coordinates of space.

### Phase 2: The Language of Perpendicularity

We are given that the perpendiculars from $A, B, C$ to the sides of the other triangle meet at a point $H$. Let the position vector of this point be $\vec{h}$.

Now, how do we speak 'perpendicular' in the language of vectors? We use the dot product. If two vectors are perpendicular, their dot product is zero.

Thus, for the altitude from $A$ to $QR$, we write:

$$(\vec{a}-\vec{h}) \cdot (\vec{q}-\vec{r}) = 0$$

This is the heartbeat of our proof. We repeat this for the other two altitudes:

$$(\vec{b}-\vec{h}) \cdot (\vec{r}-\vec{p}) = 0$$

$$(\vec{c}-\vec{h}) \cdot (\vec{p}-\vec{q}) = 0$$

Look at these three equations. They are beautiful, aren't they? They represent the constraints imposed by the concurrency at $H$.

### Phase 3: The Algebraic Magic

Here is where the intuition kicks in. We want to prove something about the *other* set of perpendiculars, but our equations are currently shackled to the point $H$.

We need to set $H$ free. What happens if we sum these three equations? Let us try:

$$\sum (\vec{a}-\vec{h}) \cdot (\vec{q}-\vec{r}) = 0$$

When we expand this, we get two distinct parts: one involving the vertices $\vec{a}, \vec{b}, \vec{c}$ and one involving the point $\vec{h}$.

$$\sum \vec{a} \cdot (\vec{q}-\vec{r}) - \vec{h} \cdot \sum (\vec{q}-\vec{r}) = 0$$

Now, look at the second term. The sum of the vectors $(\vec{q}-\vec{r}) + (\vec{r}-\vec{p}) + (\vec{p}-\vec{q})$ is exactly the zero vector $\vec{0}$.

The $\vec{h}$ term vanishes! It is like magic—the point $H$ disappears, leaving us with an invariant relation that depends only on the triangles themselves:

$$\vec{a} \cdot (\vec{q}-\vec{r}) + \vec{b} \cdot (\vec{r}-\vec{p}) + \vec{c} \cdot (\vec{p}-\vec{q}) = 0$$

### Phase 4: The Symmetry Reveal

We are almost there. We have an equation, but it is not yet in the form we need to prove the second concurrency.

Let us expand the dot products and rearrange the terms. We group them by $\vec{p}, \vec{q},$ and $\vec{r}$:

$$\vec{p} \cdot (\vec{c}-\vec{b}) + \vec{q} \cdot (\vec{a}-\vec{c}) + \vec{r} \cdot (\vec{b}-\vec{a}) = 0$$

Wait! Look closely at the terms in the parentheses. $\vec{c}-\vec{b}$ is the vector $\vec{BC}$, $\vec{a}-\vec{c}$ is $\vec{CA}$, and $\vec{b}-\vec{a}$ is $\vec{AB}$.

Our equation transforms into:

$$\vec{p} \cdot \vec{BC} + \vec{q} \cdot \vec{CA} + \vec{r} \cdot \vec{AB} = 0$$

This is the exact condition for the perpendiculars from $P, Q, R$ to the sides $BC, CA, AB$ to be concurrent. We have proven that the property is mutual.

These are what we call **Orthologic Triangles**. You have just navigated through a beautiful proof, turning a complex geometric problem into a symphony of vector algebra. Take a moment to appreciate the symmetry—the math does not lie.

Visualized Solution

Setting the Stage: Triangles $A B C$ and $P QR$

The Point of Concurrency $H$

Vector Condition for Perpendicularity

Writing the Three Equations

Summing the Equations

Expanding the Sum

Vanishing of the $h$ Term

The Invariant Relation

Expanding the Dot Products

Rearranging the Terms

Factoring the Rearranged Equation

Geometric Interpretation

The Final Concurrency

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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