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USAMO 1972 #2, #3 July 21, 2009

Posted by lumixedia in Problem-solving.
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I think I might as well just start going through the USAMOs in chronological/numerical order.

USAMO 1972 #2. A given tetrahedron {ABCD} is isosceles, that is {AB=CD}, {AC=BD}, {AD=BC}. Show that the faces of the tetrahedron are acute-angled triangles.

My (also the official) solution. By SSS congruency, we have {\triangle ABC\cong\triangle BAD\cong\triangle CDA\cong\triangle DCB}. Then {\angle BAC+\angle CAD+\angle DAB=\angle BAC+\angle ACB+\angle CBA=180^{\circ}}. But we must have {\angle BAC+\angle CAD>\angle DAB} and similarly for the other two pairs of angles at {A}, so none of {\angle BAC}, {\angle CAD}, {\angle DAB} can be {90^{\circ}} or greater. So all angles in {\triangle ABC} are acute and therefore all angles in all the triangles are acute.

The official solution just asserts that the angles at {A} satisfy the triangle inequality. This seems intuitive, but I am having an unexpected amount of trouble trying to prove it. My initial attempt to transform it into the regular triangle inequality failed; Damien suggested using vectors, which is probably a good idea, but I haven’t worked it out yet and I need to get back to my project soon. Anyway.

USAMO 1972 #3. A random number selector can only select one of the nine integers {1,2,...,9}, and it makes these selections with equal probability. Determine the probability that after {n} selections ({n>1}), the product of the {n} numbers selected will be divisible by {10}.

My (also the official) solution. After {n} selections, the total number of possible distinct sequences is {9^n}, the number of sequences not containing an even number is {5^n}, the number of sequences not containing a {5} is {8^n}, and the number of sequences containing neither an even number nor a {5} is {4^n}. So by the principle of inclusion-exclusion, the desired probability is

\displaystyle \frac{9^n-5^n-8^n+4^n}{9^n}.

That would’ve been about an AMC-12 #15 today…

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Comments»

1. Tirasan Khandhawit - July 23, 2009

I think I have an alternate solution to {\#2}.

First denote {AB, AC, AD} by vectors {x, y, z} respectively. The given conditions can be written as {\left|x-y\right| = \left|z\right|} , {\left|y-z\right| = \left|x\right|}, and {\left|z-x\right| = \left|y\right|}. Take the sum of squares of these three equations, we obtain \displaystyle  \left|x\right|^2 + \left|y\right|^2 + \left|z\right|^2 = 2(x\cdot y + y\cdot z + z\cdot x)

Suppose that {\angle BAC} is not acute, this implies {x\cdot y \leq 0}. Then {2(y\cdot z + z\cdot x) \geq 2(x\cdot y + y\cdot z + z\cdot x) = \left|x\right|^2 + \left|y\right|^2 + \left|z\right|^2}.

On the other hand, by Cauchy-Schartz and AM-GM, we have {2z\cdot (x+y) \leq 2 \left|z\right| \left|x+y\right| \leq \left|z\right|^2 + \left|x+y\right|^2} {= \left|x\right|^2 + \left|y\right|^2 + \left|z\right|^2 + 2x\cdot y \leq \left|x\right|^2 + \left|y\right|^2 + \left|z\right|^2}.

Since we have equality, the condition for Cauchy-Schwartz implies that {z} and {x+y} are parallel. This contradicts that {ABCD} is a tetrahedron. \qed

For the triangle inequality at {A}, I think I finally found a proof but it’s a little trickier than I thought. First assume that {\angle BAC} is the largest angle, then after some rigid motion we can also assume that {x = (1,0,0), y = (\cos{\alpha},\sin{\alpha},0), z = (\cos{\beta} \cos{\gamma},\sin{\beta}\cos{\gamma},\sin{\gamma})} with {0 < \alpha < \pi} and {0 < \gamma \leq \pi /2}. We want to show that \displaystyle  \alpha < \arccos{(\cos{\beta} \cos{\gamma})} + \arccos{(\cos{(\alpha - \beta)} \cos{\gamma})}

The main argument is to observe that for constants {a,b \in \left[-1,1\right]} a function {f(t) = \arccos{(a t)} + \arccos{(b t)}} is monotone on {t \in \left[-1,1\right]} and to check that {\alpha \leq \arccos{(\cos{\beta})} + \arccos{(\cos{(\alpha - \beta)})}}. The rest is case analysis.

I will fill in the detail if someone want to see. I don’t know if there is another proof, but will be very interested to see.

2. lumixedia - July 23, 2009

That’s a really neat solution to the original problem. Thanks! Thanks for the proof of the angle triangle inequality, too–I’m glad I’m not the only one who didn’t find it as obvious as the official solution-writers apparently did ^_^

The official packet also provides a Euclidean solution which I might write up in another comment to this post.

3. lumixedia - July 26, 2009

Here’s the Euclidean solution from the official packet.

Assume for the sake of contradiction that {\angle BDC} is not acute. Let {M} be the midpoint of {BC}. {AM} and {DM} are corresponding medians of congruent triangles, so {AM=DM}. Consider the circle with {BC} as a diameter. {D} must lie on or inside the circle to give {\angle BDC\ge 90^{\circ}}, so that we have {MD\le BC/2}. Then

\displaystyle AM+MD=MD+MD\le BC=AD

Which is impossible since {AM}, {MD}, {AD} are the sides of a triangle.


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