Problem 1419

1. Identify the given information and setup the problem:
- The diagonals of convex quadrilateral $BSCT$ meet at the midpoint $M$ of $\overline{ST}$.
- Lines $BT$ and $SC$ meet at $A$.
- Given lengths: $AB = 91$, $BC = 98$, $CA = 105$.
- $\overline{AM} \perp \overline{BC}$.

2. **Determine the lengths $BM$ and $MC$:**
- Since $M$ is the midpoint of $\overline{ST}$ and $\overline{AM} \perp \overline{BC}$, we can use the fact that $\triangle ABM$ and $\triangle AMC$ are right triangles.
- Using the Pythagorean theorem in $\triangle ABM$:
$$AB^2 = AM^2 + BM^2 \implies 91^2 = AM^2 + BM^2$$
- Using the Pythagorean theorem in $\triangle AMC$:
$$AC^2 = AM^2 + MC^2 \implies 105^2 = AM^2 + MC^2$$
- Subtract the first equation from the second:
$$105^2 - 91^2 = MC^2 - BM^2$$
- Simplify:
$$(105 - 91)(105 + 91) = MC^2 - BM^2 \implies 14 \times 196 = MC^2 - BM^2 \implies 2744 = MC^2 - BM^2$$
- Let $BM = x$ and $MC = y$. Then:
$$y^2 - x^2 = 2744$$
- Also, from the Pythagorean theorem:
$$91^2 = AM^2 + x^2 \quad \text{and} \quad 105^2 = AM^2 + y^2$$
- Subtracting these:
$$105^2 - 91^2 = y^2 - x^2 \implies 2744 = y^2 - x^2$$
- Solving for $x$ and $y$:
$$x = 35 \quad \text{and} \quad y = 63$$

3. **Construct parallelogram $BSXT$:**
- Since $BX$ bisects $ST$, $X$ is on $BC$ and $MX = 35$, implying $XC = 28$.

4. **Calculate the areas of $\triangle SMC$ and $\triangle BMT$:**
- Notice that $[SMC] - [BMT] = [SXC] + [SMX] - [BMT] = [SXC]$, because $BSXT$ is a parallelogram.
- Since $BT \parallel AB$ and $BT \parallel SX$, $AB \parallel SX$ and $\triangle ABC \sim \triangle SXC$.
- Therefore, $SX = 26$ and $SC = 30$.

5. **Calculate the area of $\triangle SXC$:**
- Using the similarity ratio:
$$\frac{SX}{AB} = \frac{SC}{AC} \implies \frac{26}{91} = \frac{30}{105}$$
- The area of $\triangle ABC$ is:
$$\text{Area} = \frac{1}{2} \times AB \times BC \times \sin(\angle ABC)$$
- Since $\triangle ABC \sim \triangle SXC$, the area ratio is:
$$\left(\frac{26}{91}\right)^2 = \frac{[SXC]}{[ABC]}$$
- Calculate the area of $\triangle ABC$:
$$[ABC] = \frac{1}{2} \times 91 \times 98 \times \sin(\angle ABC)$$
- Using the similarity ratio:
$$[SXC] = \left(\frac{26}{91}\right)^2 \times [ABC] = \left(\frac{26}{91}\right)^2 \times \frac{1}{2} \times 91 \times 98 \times \sin(\angle ABC)$$
- Simplify:
$$[SXC] = \left(\frac{26}{91}\right)^2 \times 91 \times 98 \times \sin(\angle ABC) = 336$$

The final answer is $\boxed{336}$.