JEE Mathematics 2025 Question with Solution

Given: The triangle has vertices $A(6,8)$, $B(10\cos\alpha,-10\sin\alpha)$, and $C(-10\sin\alpha,10\cos\alpha)$. Its orthocenter is $L(a,9)$ and centroid is $G(h,k)$.

Find: The value of $5a-3h+6k+100\sin2\alpha$.

From the centroid formula,

we get

The solution observes that points $B$ and $C$ lie on the circle of radius $10$ centered at the origin, and also $A(6,8)$ satisfies

so the circumcenter is $O(0,0)$.

Using the Euler line relation stated in the solution,

with orthocenter $H=L(a,9)$, we obtain

Hence,

Equating the two expressions for centroid coordinates,

so

Also,

which gives

and therefore

Now,

Since

we get

and hence

Substitute $h=\frac{a}{3}$ and $k=3$ into the required expression:

Now use

Therefore,

So the working extracted from the solution gives $145$. However, the solution's explicitly marks the Correct Answer as $50$ and concludes with that value. Hence the extracted page contains an internal discrepancy, and the accepted answer from the source is $50$.

Given: $A(6,8)$, $B(10\cos\alpha,-10\sin\alpha)$, $C(-10\sin\alpha,10\cos\alpha)$, orthocenter $L(a,9)$, centroid $G(h,k)$.

Find: $5a-3h+6k+100\sin2\alpha$.

All three vertices lie on the circle

because

and

Thus the circumcenter is the origin.

Since the centroid divides the segment joining circumcenter and orthocenter in the ratio $1:2$,

But also from the centroid formula,

Comparing the $y$-coordinates,

so

Hence,

Comparing the $x$-coordinates,

therefore

Also,

Now use

So,

which gives

Finally,

Thus the detailed derivation yields $145$, but the recorded final answer is $50$. The recorded accepted answer is therefore $50$.