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Source: moldova

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#1 h Apr 30, 2005, 8:58 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Let $ABCD$ be a square, and $C$ the circle whose diameter is $AB.$ Let $Q$ be an arbitrary point on the segment $CD.$ We know that $QA$ meets $C$ on $E$ and $QB$ meets it on $F.$ Also $CF$ and $DE$ intersect in $M.$ show that $M$ belongs to $C.$

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Sailor

#2 h May 1, 2005, 2:57 PM • 2 Y

Y by Adventure10, Mango247

Very strange, that it's from Moldova since I don't remember it.
Anyway the first thing that comes in my mind is an analitic solution, it's really not hard to compute the equation of the two intersecting lines(CF and DE), and you just check that it belong to C(here you may choose the origin the midpoint of $AB$ )!

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#3 h May 1, 2005, 3:57 PM • 2 Y

Y by Adventure10, Mango247

believe me, an analityc solution can't be done by a human,i ve already tried , i am not sure of the source(maybe bosnia) but it remains a very hard problem, much more than it looks . can someone have a try?

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#4 h May 1, 2005, 9:03 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

Let BE and AD intersect at S, and let AF and BC intersect at T, and define $\alpha=\angle BAQ$ and $\beta=\angle QBA$ . Also let Y be the center of the square. Since $\angle SBA = 90-\angle BAE = \angle QAD$ , and since AB = DA, right triangles QDA and SAB are congruent. This means that S is the image of Q when rotated 90 degrees around Y. By similar arguments, T is the image of Q when rotated around Y 90 degrees the other way. So TQS is an isosceles right triangle. Let $\angle QSD = \delta$ , so that $\angle CTQ = 90-\delta$ .

Let's calculate angle CMD. First, since QDSE is cyclic, $\angle MDC = \angle ESQ = 180-\angle ASB-\angle QSD = 180-\alpha-\delta$ . (We have used the fact that $\triangle ASB\sim\triangle EAB$ which implies $\angle ASB = \angle EAB = \alpha$ .) Likewise, $\angle DCM = \angle QTF = 180-\beta-(90-\delta)$ . We then have $\angle CMD = 180-\angle MDC - \angle DCM$ , and after some calculation this comes out to $\alpha+\beta-90$ .

Now lets calculate angle FBE. We have $\angle FBA = \beta$ and $\angle EBA = 90-\alpha$ , so $\angle FBE = \beta-(90-\alpha)=\alpha+\beta-90$ .

Since we have $\angle FBE = \alpha+\beta-90 = \angle FME$ , EFBM is cyclic, so M is on the circle as desired.

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Sailor

#5 h May 2, 2005, 1:47 PM • 2 Y

Y by Adventure10, Mango247

wisemaniac wrote:

believe me, an analityc solution can't be done by a human,i ve already tried , i am not sure of the source(maybe bosnia) but it remains a very hard problem, much more than it looks . can someone have a try?

Well it took ''only'' 2 pages for me to write the solution, so there is a human analitic approach!

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#6 h May 4, 2005, 3:42 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

a really beautiful problem as it is a special case of a special case you get from a special case

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so, the general problem is the following:

suppose we have a cubic $E$ in the plane and points $A,B,C,D$ on it s.t. the line $AC$ and $BD$ meet on $E$ . let $X$ be an arbritary point of $E$ . let $P$ be the third point of intersection of $XA$ with $E$ and let Q be the third points of intersection of $XB$ with $E$ . then $PD$ and $QC$ concur on $E$ .

proof: i will do the proof in the language of divisors and linear equivalence; it's quite the same as the language of addition of points on cubics. so we know that $A+C\sim B+D$ by the first condition. furthermore we know that $P+X+A\sim P+R_1+D$ where $R_1$ is the third point of intersection of $PD$ with $E$ ; this is clear since both sides are the zeroes of some linear functions(the intersections with lines). Thus we have $R_1\sim A-D+X$ . defining $R_2$ analogously, we get $R_2\sim B-C+X$ . Thus we have, since $A-D\sim B-C$ that $R_1\sim R_2$ and this means that $R_1=R_2$ (this is the crucial step; this implication holds for all curves of genus at least 1).

now for our problem we take $E$ as the union of the $C$ and the line $CD$ . it is clear that $AC$ and $BD$ meet on $E$ . therefore we immediately get the result.

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#7 h Jul 1, 2005, 10:55 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

The problem is a one-liner using Pascal, if one lets the center O of the square ABCD join the picture. Here is an extended version of the problem:

Extended problem. Let ABCD be a square, and k the circle with diameter AB. Let Q be an arbitrary point on the segment CD. Let the line QA meet the circle k at a point E (apart from A), and let the line QB meet the circle k at a point F (apart from B). Furthermore, let O be the center of the square ABCD, and let R be the orthogonal projection of the point Q on the line AB. Then, prove that the lines CF, DE and OR intersect at one point which lies on the circle k.

Solution of the extended problem. The center O of the square ABCD clearly must lie on both diagonals AC and BD of the square and on the circle k with diameter AB (since < AOB = 90°). Now, let the line OR meet the circle k at a point M (apart from O). We will show that this point M lies on the lines CF and DE; then, of course, the problem will be solved.

Since the points Q and R lie on the sides CD and AB of the square ABCD, respectively, and satisfy $QR\perp AB$ , they are symmetric to each other with respect to the parallel to the side AB of the square through its center O. Of course, the diagoanls AC and BD are also symmetric to each other with respect to this parallel. Since the point O lies on this parallel, it thus follows from symmetry that < QOC = < ROB.

Let the line OQ meet the circle k at a point N (apart from O). Then, < QOC = < ROB becomes < AON = < MOB. But since the points A, O, N, B lie on one circle (namely, on the circle k), we have < AON = < ABN, and since the points M, O, B, N lie on one circle (namely, on the circle k), we have < MOB = < MNB. Thus, < ABN = < MNB, and hence NM || AB. But since ABCD is a square, we have AB || CD. Thus, the lines NM, AB and CD are all parallel to each other; hence, they concur at one point (namely, at an infinite point). In other words, the point of intersection $AB\cap NM$ lies on the line CD.

The point of intersection $EA\cap ON$ is the point Q and also lies on the line CD.

Now, applying the Pascal theorem to the cyclic hexagon EABONM, which is inscribed in the circle k, we see that the points of intersection $EA\cap ON$ , $AB\cap NM$ and $BO\cap ME$ are collinear. Since the points $EA\cap ON$ and $AB\cap NM$ both lie on the line CD, it follows that the third point $BO\cap ME$ must also lie on the line CD. In other words, the lines BO, ME and CD concur. But we know that the lines BO and CD intersect each other at the point D; thus, the line ME passes through the point D. In other words, the point M lies on the line DE. Similarly, the point M lies on the line CF. Thus, the problem is solved.

EDIT: We can show a bit more: We have $\frac{AM}{BM}=\frac{DQ}{CQ}$ , where M is the point of intersection of the lines CF, DE and OR.

Proof. Since the points B, O, A and M all lie on the circle k, we have < BMO = < BAO. But since ABCD is a square and O is its center, < BAO = 45°. Thus, < BMO = 45°. Similarly, < AMO = 45°. Hence, < AMO = < BMO, so that the line MO is the angle bisector of the angle AMB. Now, in a triangle, the angle bisector of an angle divides the opposite side in the ratio of the adjacent sides. Hence, in the triangle AMB with the angle bisector MO of its angle AMB, we have $\frac{AR}{BR}=\frac{AM}{BM}$ . But since the quadrilateral BRQC is a rectangle (in fact, it has three right angles: < RBC = < ABC = 90°, < BCQ = < BCD = 90° and < QRB = 90°), we have BR = CQ, and similarly AR = DQ, and thus this equation becomes $\frac{DQ}{CQ}=\frac{AM}{BM}$ . This completes our proof.

Darij

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jayme

#8 h Dec 1, 2010, 10:39 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
an article concerning "Geometric miniatures on a square" with original proofs has been put on my website
http://perso.orange.fr/jl.ayme vol. 7 Miniatures géométriques sur un carré p. 10
Sincerely
Jean-Louis

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