An adventitious angle problem concerning

Transcription

1 n adventitious angle problem concerning and 7 / Darij Grinberg The purpose of this note is to give two solutions of the following problem (Fig 1): Let be an isosceles triangle with and 1 Let be a point on the side of this triangle which satisfies 1 rove that holds if and only if 7 Fig 1 It is not hard to solve this problem using trigonometry or complex numbers (see, e g, the MathLinks discussion 89 for the direction 7 ) Here, we will present two synthetic solutions of the problem; the first one was given (for the direction ) by Stefan V (a pseudonym), the second one is apparently original 7 First solution (Stefan V): efore solving the problem, we recall two facts on parallelograms The first one is a pretty well-known formula: Lemma 1 Let D be a parallelogram Then, D In other words, the sum of the squares of the diagonals of a parallelogram is equal to the double sum of the squares of two adjacent sides (See Fig )

2 Fig Lemma 1 is most easily proven using vectors and their scalar products: Since D is a parallelogram, we have D, or, equivalently, D Thus D D, and hence D D, so Lemma 1 is proven Note that Lemma 1 is more known in the form D D D, which is trivially equivalent to D since D and D (because D is a parallelogram) The next property of parallelograms applied below will be: Lemma Let D be a parallelogram If, then D In other words, if in a parallelogram, a diagonal is times as long as a side, then the other diagonal is times as long as the other side (See Fig 3) D D Fig 3 In fact, Lemma is a trivial corollary of Lemma 1: If, then ; subtracting this from the equation D which holds by Lemma 1, we obtain D, so that D, and Lemma is proven

3 X Fig There is also an alternative proof of Lemma using similar triangles (Fig ): Let X be the reflection of the point in the point Then, X On the other hand, D, since D is a parallelogram Thus, X D Together with X D (what follows from D, what is because D is a parallelogram), this yields that the quadrilateral DX is a parallelogram, so that X D Now, we supposed that, so that In other words, ut X yields X X, so this becomes X Since we also trivially have X, we can conclude that the triangles and X are similar Thus, X Since X D and, this becomes D D This again proves Lemma Now we come to the actual solution of the problem: In order to solve the problem, we have to prove two assertions: ssertion 1: If, then 7 ssertion : If 7, then ; hence, efore we verify these two assertions, we perform some observations independent of the validity of and 7 (See Fig 5) Let the parallel to the line through the point meet the parallels to the lines and through the point at the points S and R We have S, or, equivalently, S, and we have S ; thus, the quadrilateral S is a parallelogram Thus, S On the other hand, we have R, or, equivalently, R, and we have R ; thus, the quadrilateral R is a parallelogram This yields R Hence, RS R S Together with RS this implies that the quadrilateral RS is a parallelogram Let Since triangle is isosceles, its base angle then equals D

4 Since R, we have R, so that R Now R 1 ; thus, the triangle R is isosceles, so its base angle is R R Hence, R R 3 Now, R implies R, so that R Thus, the sum of angles in triangle R yields R R R 3 3 Now, we have R if and only if the triangle R is isosceles with base R; this holds if and only if R R, i e if 3 ; but this is obviously equivalent to 3, hence to 7 So we have shown that R holds if and only if 7

5 S R Fig 5 Now, we will prove the ssertions 1 and We start with the proof of ssertion 1: ssume that Since 1, this rewrites as y Lemma, applied to the parallelogram R, this entails R Since, this rewrites as R ccording to Lemma, applied to the parallelogram RS, this leads to S R ut since 1, we can rewrite the equation in the form as well, and thus, from Lemma, applied to the parallelogram S, we conclude that S omparing this with S R, we get R s showed above, this is equivalent to 7, i e to 7, and thus ssertion 1 is proven More difficult is the proof of ssertion : ssume that 7 In other words, 7 ccording to the above, this yields R pplication of Lemma 1 to the parallelogram S yields S, what, in view of 1 1, becomes S 1 pplication of Lemma 1 to the parallelogram R yields R, what, in view of and 1 1, becomes R 1 pplication of Lemma 1 to the parallelogram RS yields R S R, what, in view of R, becomes R S Thus,

6 so that S R R S 1 1, Thus, ssertion is proven, and the solution of the problem is complete Second solution: Q Fig 6 (See Fig 6) The point lies on the side of triangle and satisfies 1 Let Q be the point on the side of triangle satisfying Q 1 Since the triangle is isosceles with, from symmetry it then follows that Q, Q and Q Since Q, we have Q Since triangle is isosceles with, we have Thus Q Q Thus, the quadrilateral Q is cyclic, so the tolemy theorem yields Q Q Q Since 1, Q and Q, this becomes 1 Q, what simplifies to Q The triangle is isosceles with the base ; let be its base angle Then, the angle at its apex is onsequently,

7 M Q N Fig 7 (See Fig 7) Now let M be the point on the ray Q satisfying MQ Since Q, we have QM, thus QM, and thus MQ QM ; hence, the triangle MQ is isosceles with base Q, and it has the same base angle as the isosceles triangle (in fact, the base angle of triangle is, too) Furthermore, it has the same base as triangle (since Q 1 and 1) Hence, the isosceles triangle MQ is congruent to the isosceles triangle Therefore, the legs of these two triangles are equal: QM Since and MQ, we have M MQ 3

8 M U Q V N (See Fig 8) Let the angle bisector of the angle M intersect the line Q at a point U Then, U M 3 onsequently, UQ U 3 3 On the other hand, QU QM ; by the sum of angles in triangle UQ, we thus have UQ QU UQ UQ Therefore, the triangle UQ is isosceles with QU Q Since Q 1, this means that QU 1 Together with QM, this leads to MU QM QU 1 Similarly to the point M on the ray Q satisfying MQ, we can construct a point N on the ray Q satisfying N Similarly to the point U, we then define the point of intersection V of the angle bisector of the angle QN with the line Q Similarly to the above equation QU 1, we can now prove that V 1 s showed above, UQ In other words, UV Similarly, VU On the other hand, UQ and UQ Thus, UV UQ and VU UQ Hence, the triangles UV and QU are similar; this yields U : UV Q : U, so that U Q UV Since Q Q 1, this becomes U UV Now, UV QU QV QU V Q 1 1 Q Q, and hence Q Q UV MU U (since MU and U UV) s the triangle MQ is congruent to the triangle, we have QM ; in other words,

9 UM On the other hand, the line U is the angle bisector of the angle M, and this yields MU M Now, we have if and only if ut since MU U, we have if and only if MU U, thus if and only if MU U, i e if and only if the triangle MU is isosceles with base M This, in turn, is equivalent to the equality of its angles UM and MU; but because of UM and MU 3, these angles are equal if and only if 3 This simplifies to 3, and thus to 7 ombining, we see that if and only if ; hence the problem is solved 7