Math Olympiad Problems for Grades 9 and 10: Very Difficult Problems with Solutions

Solution:
We factor by adding a constant to both sides (Simon’s Favorite Factoring Trick).
[tex]xy - 3x - 3y = 17 \Rightarrow x(y - 3) - 3y = 17[/tex].
Add [tex]9[/tex] to both sides to complete the factorization:
[tex]x(y - 3) - 3(y - 3) = 17 + 9 \Rightarrow (x - 3)(y - 3) = 26[/tex].
The integer factors of [tex]26[/tex] are [tex]1, 2, 13, 26[/tex] (and their negatives).
To maximize [tex]x[/tex], we set [tex]x - 3[/tex] to the largest positive factor of [tex]26[/tex]: [tex]x - 3 = 26 \Rightarrow x = 29[/tex].

Solution:
Let [tex]R[/tex] be the radius of the quarter-circle and [tex]r[/tex] the radius of the inscribed circle.
The distance from the corner to the inscribed circle’s center is [tex]r\sqrt{2}[/tex].
Extending this to the outer edge gives
[tex]R = r\sqrt{2} + r = r(\sqrt{2} + 1)[/tex].
Squaring both sides: [tex]R^2 = r^2(3 + 2\sqrt{2})[/tex].
The area of the quarter-circle is [tex]\frac{1}{4}\pi R^2 = 36\pi[/tex], so [tex]R^2 = 144[/tex].
Thus [tex]r^2 = \frac{144}{3 + 2\sqrt{2}} = 144(3 - 2\sqrt{2})[/tex].
Area of inscribed circle [tex]= \pi r^2 = 144(3 - 2\sqrt{2})\pi \text{ cm}^2[/tex] (approximately [tex]77.6 \text{ cm}^2[/tex]).

Solution:
Use variable substitution: let [tex]x = \sqrt{a}[/tex] and [tex]y = \sqrt{b}[/tex].
The system becomes [tex]x^2y + y^2x = 30[/tex] and [tex]x + y = 5[/tex].
Factor the first equation: [tex]xy(x + y) = 30[/tex].
Substitute the second equation into this:
[tex]xy(5) = 30 \Rightarrow xy = 6[/tex].
The product [tex]a \times b = (\sqrt{a} \times \sqrt{b})^2 = (xy)^2 = 6^2 = 36[/tex].

Solution:
Using the Power of a Point theorem for an external point [tex]P[/tex]: [tex]PA \cdot PB = PC \cdot PD[/tex].
For secant [tex]PAB[/tex]: [tex]PB = 6[/tex] and [tex]PA = PB + AB = 6 + 18 = 24[/tex].
Power [tex]= 24 \times 6 = 144[/tex].
For secant [tex]PCD[/tex]: let [tex]PC = x[/tex].
Then [tex]PD = x + 10[/tex].
Equating the powers: [tex]x(x + 10) = 144 \Rightarrow x^2 + 10x - 144 = 0[/tex].
Factoring gives [tex](x + 18)(x - 8) = 0[/tex].
Since distance must be positive, [tex]x = 8[/tex] cm.

Solution:
If [tex]C[/tex] is assigned Core [tex]2[/tex], then [tex]B[/tex] must be assigned Core [tex]3[/tex] (since [tex]B[/tex] can only be [tex]2[/tex] or [tex]3[/tex]).
Core [tex]5[/tex] cannot be [tex]D[/tex] or [tex]C[/tex] (and [tex]C[/tex] is already [tex]2[/tex]), so [tex]D[/tex] is restricted from [tex]5[/tex].
Now check the options for Core [tex]1[/tex] ([tex]A[/tex] or [tex]E[/tex]): Case 1: Core [tex]1[/tex] is [tex]A[/tex].
The remaining cores are [tex]4[/tex] and [tex]5[/tex] for [tex]D[/tex] and [tex]E[/tex].
Since [tex]D[/tex] cannot be [tex]5[/tex], [tex]D[/tex] must be [tex]4[/tex], and [tex]E[/tex] must be [tex]5[/tex].
Case 2: Core [tex]1[/tex] is [tex]E[/tex].
By rule 4, [tex]D[/tex] must be Core [tex]4[/tex].
[tex]A[/tex] must be [tex]5[/tex].
In both possible scenarios, [tex]D[/tex] must be assigned Core [tex]4[/tex].