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Problem: The length of diameter AB is a two digit integer. Reversing the digits gives the length of a perpendicular chord CD.The distance from their intersection point H to the center O is a positive rational number. Determine the length of AB. My attempt: Let the length of AB be 10a+b. Then the length of CD is 10b+a. Now since AO is a radius, H bisects CD. Therefore, HD=CD/2=(10b+a)/2 and OD=AO=AB/2=(10a+b)/2. In triangle OHD, $OH^2+HD^2=OD^2$, $OH^2=OD^2-HD^2$

$OH^2=(\frac{10a+b}{2})^2-(\frac{10b+a}{2})^2$

$=>OH^2=\frac{(10a+b)^2}{4}-\frac{(10b+a)^2}{4}$

$=>OH^2=\frac{100a^2+b^2+20ab-100b^2-a^2-20ab}{4}$

$=>OH^2=\frac{99a^2-99b^2}{4}$

$=>OH^2=\frac{99}{4}*(a^2-b^2)$

$=>OH=\frac{3\sqrt{11}}{2}*\sqrt{(a^2-b^2)}$

Since it is given that OH is a positive rational number, $\sqrt{(a^2-b^2)}$ should be of the form $x\sqrt{11}$ where $x$ is an integer. Therefore $a^2-b^2$ must be of the form $11x^2$. $a^2-b^2=(a-b)(a+b)$. Since $0<a,b\leq 9$ and both cannot be $0$, we have only two cases when $a^2-b^2$ is a multiple of $11$: 1)$a-b=0$ and 2) $a+b=11$ If the first case is true, OH=0-which is not possible because OH is a positive rational number. Therefore $a+b=11$. Considering all a,b such that $0\leq a,b \leq 9$ and such that $a+b=11$, the only a,b which makes $a-b$ a perfect square is $(6,5)$. Hence the length of $AB=10a+b=65$.

Is my attempt to the problem correct?

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  • $\begingroup$ Sound logical enough. $\endgroup$ – Mick Sep 4 '16 at 14:03
  • $\begingroup$ Where did you get these questions? $\endgroup$ – N.S.JOHN Sep 4 '16 at 14:40
  • $\begingroup$ I was asked these questions in Junior Math Olympiad conducted by our school in India $\endgroup$ – Gayatri Sep 5 '16 at 7:28
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Yes, your attempt to the problem is correct. here is another way of solving it.

The equation of a circle is $$x^2 + y^2 = R^2\Rightarrow x^2 + y^2 = (\frac{d}2)^2$$ where, R and d are the radius and the diameter of the circle respectively.
So, the equation for the top semi circle is $$y = \sqrt{(\frac{d}2)^2 - x^2}$$ Let, the diameter of the circle be of the form $$d = 10a + b$$ where, $a$ and $b$ are whole numbers between 0 and 9 inclusive. So, $a$ and $b$ are the digits of the two digit number.
Also, it is given that the length of cord CD is obtained by reversing the digits of the diameter. So, $$CD = 10b + a$$ Let us define a function $C(x)$ which gives us the length of a perpendicular chord, whose input is the distance of the center of the chord from the center of the circle.

We know that the length of half of the perpendicular chord is given by $$\sqrt{(\frac{d}2)^2 - x^2}$$ So,$$C(x) = 2\sqrt{(\frac{d}2)^2 - x^2}\Rightarrow 2\sqrt{(\frac{10a + b}2)^2 - x^2}$$ Now, let the distance of CD from the center of the circle be a rational number of the form $\frac{p}q$, where $p$ and $q$ are relatively prime integers.

Hence, the length of CD is given by $$C(\frac{p}q) = 2\sqrt{(\frac{10a + b}2)^2 - (\frac{p}q)^2}$$$$\Rightarrow10b + a = 2\sqrt{(\frac{10a + b}2)^2 - (\frac{p}q)^2}$$ Simplifying the equation will give us,$$(\frac{2p}{3q})^2 = 11(a^2-b^2)$$Since, the LHS is a perfect square, RHS must also be a perfect square. As, the RHS is already a multiple of $11$, $ a^2-b^2$ must be an odd power of 11, like $11^1, 11^2, 11^3$, etc.
Since, $a$ and $b$ are whole numbers between 0 and 9, the maximum value of $a^2-b^2$ will be when $a = 9$ and $b = 0$ , i.e. $81$.

Because, all odd powers of $11$ are greater than $81$ except $11$ itself, $a^2 - b^2$ must be $11$.
So, when we see a list of perfect squares from $1^2$ to $9^2$ $$1, 4, 9, 16, 25, 36, 49, 64, 81$$we see that only $36\;(i.e.\;6^2)$ and $25\;(i.e.\;5^2)$ are $11$ apart. Hence, we can choose, $a = 6$ and $b = 5$, to satisfy the equation $a^2 - b^2 = 11$.

Therefore, the length of the diameter of the circle is $65$ units.

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