Question

Let $PQRS$ be a quadrilateral. Two circles $O_1$ and $O_2$ are inscribed in triangles $PQR$ and $PSR$ respectively. Circle $O_1$ touches $PR$ at $M$ and circle $O_2$ touches $PR$ at $N$. Find the length of $MN$. % Statement I A circle is inscribed in the quadrilateral $PQRS$. % Statement II The radii of the circles $O_1$ and $O_2$ are 5 and 6 units respectively.

• A. Statement I alone is sufficient to answer the question.
• B. Statement II alone is sufficient to answer the question.
• C. Statement I and II together are sufficient, but neither alone is sufficient.
• D. Either Statement I or Statement II alone is sufficient.
• E. Neither Statement I nor Statement II is necessary to answer the question.

Hint:

In Data Sufficiency problems involving circles, always check tangent equalities. For quadrilaterals, remember: a circle can be inscribed if and only if the sum of opposite sides are equal. This condition often forces equalities that directly determine the required length.

Answer 1

The Correct Option is A

Step 1: Recall tangent property.
For any circle inscribed in a polygon, tangents drawn from the same external point are equal. For example, if a circle touches sides $AB$ and $AC$ of $\triangle ABC$ at points $X$ and $Y$, then \[ AX = AY \] This equality is the foundation for solving inscribed-circle problems.

Step 2: Apply to quadrilateral $PQRS$.
Suppose quadrilateral $PQRS$ has an incircle. Then tangents from each vertex are equal: - From $P$: tangent lengths $a$ and $d$ - From $Q$: tangent lengths $b$ and $a$ - From $R$: tangent lengths $c$ and $b$ - From $S$: tangent lengths $d$ and $c$ So side lengths can be expressed in terms of tangent lengths: \[ PQ = a+b, \quad QR = b+c, \quad RS = c+d, \quad SP = d+a \]

Step 3: Condition for a quadrilateral to have an incircle.
For a circle to be inscribed inside a quadrilateral, the following must hold: \[ PQ + RS = QR + SP \] This is a standard necessary and sufficient condition.

Step 4: Relation between $M$ and $N$.
In our problem, circles $O_1$ and $O_2$ are incircles of $\triangle PQR$ and $\triangle PSR$ respectively. Let them touch $PR$ at $M$ and $N$. We need to find $MN$. From tangent property: - In $\triangle PQR$, if circle $O_1$ touches $PR$ at $M$, then \[ PM = a, \quad MR = c \] - In $\triangle PSR$, if circle $O_2$ touches $PR$ at $N$, then \[ PN = d, \quad NR = c \] Thus, \[ MN = |PM - PN| = |a - d| \]

Step 5: Use Statement I.
From the condition of an inscribed circle in quadrilateral $PQRS$: \[ PQ + RS = QR + SP \] Substituting tangent forms: \[ (a+b) + (c+d) = (b+c) + (d+a) \] Simplify: \[ a+b+c+d = a+b+c+d \] which holds true automatically. But the key observation is: if $PQRS$ has an incircle, then tangents from $P$ to sides $PQ$ and $PS$ must be equal. That means $a=d$. Therefore: \[ MN = |a-d| = 0 \] So Statement I

directly determines $MN$.

Step 6: Use Statement II.
Statement II only gives the radii of the incircles of two separate triangles: \[ r(O_1) = 5, \quad r(O_2) = 6 \] But radii alone do not fix the positions $M$ and $N$ uniquely along diagonal $PR$. The same radii can occur with different placements of the quadrilateral, leading to multiple possible $MN$ values. Hence, Statement II is not sufficient.

Step 7: Combine.
- Statement I alone $\Rightarrow MN = 0$ (sufficient). - Statement II alone $\Rightarrow$ insufficient. - Both together are unnecessary, since I alone solves it.

Final Answer: \[ \boxed{\text{A) Statement I alone is sufficient to answer the question.}} \]