JEE Mathematics 2024 Question with Solution

Given: $A(3,-1)$, $B(-2,3)$ and orthocenter $P(1,1)$ of triangle $ABC$.

Find: The value of $(\alpha+\beta)+2(h+k)$, where $C=(\alpha,\beta)$ and $(h,k)$ is the circumcenter of triangle $PAB$.

Since $P$ is the orthocenter, $CP \perp AB$ and $AP \perp BC$.

Slope of $AB$ is

$$\frac{3-(-1)}{-2-3}=\frac{4}{-5}=-\frac{4}{5}$$

Hence slope of $CP$ is

$$\frac{5}{4}$$

So the equation of line $PC$ is

$$y-1=\frac{5}{4}(x-1)$$

Also, slope of $AP$ is

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

Hence slope of $BC$ is $1$. Since $B(-2,3)$ lies on $BC$, its equation is

$$y-3=x+2$$

Now solve the two equations:

$$y=x+5$$

and

$$y-1=\frac{5}{4}(x-1)$$

Substituting,

$$x+4=\frac{5}{4}(x-1)$$ $$4x+16=5x-5$$ $$x=21$$

Then

$$y=x+5=26$$

So

$$\alpha+\beta=21+26=47$$

Now find the circumcenter of triangle $PAB$ using perpendicular bisectors.

Midpoint of $AP$ is

$$\left(\frac{3+1}{2},\frac{-1+1}{2}\right)=(2,0)$$

Slope of $AP$ is $-1$, so the perpendicular bisector has slope $1$. Therefore,

$$y=x-2$$

Midpoint of $AB$ is

$$\left(\frac{3+(-2)}{2},\frac{-1+3}{2}\right)=\left(\frac{1}{2},1\right)$$

Since slope of $AB$ is $-\frac{4}{5}$, the perpendicular bisector has slope $\frac{5}{4}$. Hence,

$$y-1=\frac{5}{4}\left(x-\frac{1}{2}\right)$$

Solving these two equations gives

$$h=-\frac{19}{2}, \quad k=-\frac{23}{2}$$

Therefore,

$$2(h+k)=2\left(-\frac{19}{2}-\frac{23}{2}\right)=-42$$

Finally,

$$(\alpha+\beta)+2(h+k)=47-42=5$$

Therefore, the correct option is C.

Given: Orthocenter $P(1,1)$ and vertices $A(3,-1)$, $B(-2,3)$.

Find: $(\alpha+\beta)+2(h+k)$.

The key observation is that the orthocenter lies on every altitude.

1. Because $CP \perp AB$, first compute slope of $AB$:

$$m_{AB}=\frac{3-(-1)}{-2-3}=-\frac{4}{5}$$

So,

$$m_{CP}=\frac{5}{4}$$

Thus,

$$y-1=\frac{5}{4}(x-1)$$

2. Because $AP \perp BC$, compute slope of $AP$:

$$m_{AP}=\frac{1-(-1)}{1-3}=-1$$

Hence,

$$m_{BC}=1$$

Through $B(-2,3)$, line $BC$ is

$$y-3=x+2$$

that is,

$$y=x+5$$

3. Point $C$ is the intersection of these two lines:

$$x+4=\frac{5}{4}(x-1)$$ $$4x+16=5x-5$$ $$x=21, \quad y=26$$

So,

$$C=(21,26)$$

and

$$\alpha+\beta=47$$

4. For circumcenter of triangle $PAB$, intersect perpendicular bisectors of $AP$ and $AB$.

Perpendicular bisector of $AP$:

• midpoint $(2,0)$
• perpendicular slope $1$

$$y=x-2$$

Perpendicular bisector of $AB$:

• midpoint $\left(\frac12,1\right)$
• perpendicular slope $\frac54$

$$y-1=\frac54\left(x-\frac12\right)$$

5. Solving,

$$h=-\frac{19}{2}, \quad k=-\frac{23}{2}$$

Therefore,

$$2(h+k)=-42$$

Hence,

$$(\alpha+\beta)+2(h+k)=47-42=5$$

So the correct option is C.