Germany Math Olympiad Problem | A Very Nice Geometry Challenge

[@shadrana1]

At 0:53, AP=PD=3,DQ=4,QC=2. AB=BC=CD=DA=6. Construction; (1) Draw a horizontal line EO, EO=x, ...............(1), (2) Draw a vertical line GO, GO=AE=y...........(2), Triangles ADQ and AEO are similar (angles are the same). AE/EQ=y/x=6/4=3/2 y/x=3/2 y=3x/2...............(3), Triangles ABP and BGO are similar (angles are the same). BG/GO=AB/AP=6/3=2/1, AG=x,GB=(6-x) and y=GO BG/GO=2/1 (6-x)/y=2/1 2y= 6-x, 3x/2=(6-x)/2 from (3), 3x=6-x 6=4x x=3/2..............(4) y=3x/2=3/2*3/2= 9/4............(5), from (3). IAOBI=y*AB/2=(9/4)*(6/2)= 27/4 sqare units. IADQI= AD*DQ/2= (6*4)/2=12 square units. IABCDI =IAOBI+IADQI+IOBCQI 6*6 =27/4 +12 +IOBCQI 24 =27/4 + IOBCQI IOBCQI= (96-27)/4 =69/4 square units. Thanks for the puzzle professor.

[@zawatsky]

ΔABP составляет четверть квадрата ABCD, т. е. 9. ΔADC составляет треть, т. е. 12. Всё вместе это 21. Осталось посчитать площадь пересечения и вычесть, потом результат вычтем уже из квадрата. Проводим горизонтально из т. P в т. F на AQ. Получаем от средней линии ΔADQ ΔFAP. Его площадь 2*3/2=3. Т. О образует от пересечения BP и AF в таком случае ΔFOP~ΔABO. Их общая высота проходит через т. О, а коэффициент подобия AB/PF=6/2=3. Мы знаем эту высоту - она равна AP и составляет 3, при этом составлена отрезками в пропорции 1:3, а значит - высота ΔFOP составляет четверть, т. е. h=¼*3=¾. S(FOP)=2*¾/2=¾. Отнимем этот маленький кусочек от S(FAP): 3-¾=2¼ - площадь пересечения нашли. Вычитаем из общей площади треугольников: 21-2¼=19-¼=18¾. Общая площадь квадрата 36=18*2. Получается, что ненамного меньше половины (чуть менее, чем наполовину ☝😉): 18-¾=17¼.

[@kateknowles8055]

Thank you Math Booster for these geometry challenges AP=PD = (4+2)/2 = 3 Angle ABP = alpha = arctan (3/6) [APB] = ½3×6 = 9 It is [ABCD]/4 [PQD] = ½3×4 = It is [ABCD] /6 Now [OPQ] could be the key to finding [BCQO] 9 + 6 + ?[OPQ] + ?[BCQO] = 36 *** I think it may be worth joining D to the midpoint (M) of BC. DM is equal in length with, and parallel to PB. Angle CDM is also alpha. Letting N be the point where DM crosses OQ, as angle (QAB) equals angle (AQD) = beta , there are similar triangles NQD and OAB and their areas have ratio 4×4 to 6x6, which is 4:9. beta = arctan (6/4) Now PAB and MCD are congruent with areas of 9 each. So the parallelogram |PBMD] has an area of 18. The angles at O are not right angles because 3:3 and 2:4 are different, that is te glitch. tan ( alpha +beta) =( tan (alpha ) + tan (beta) ) / ( 1 - tan (alpha)tan(beta)) calling this tan (gamma) or tan(g) and evaluating this (1/2 +3/2) / ( 1 - 1/2(3/2)) = 2/(1-3/4) = 8 Now tan(g) = tan (BOQ)=8 1+ tan^2 = sec^2 so( sec (g) )^2 = 64+1 so cos (g) = 1 /(65 )^½ and sin (g) = 8/((65)^½) Sine rule, triangle AOP sine(g)/AP = sine(90-alpha)/OA = sine( 90-beta)/OP which we could whizz through with a calculator to find OP OQ = AQ-OA and we could Pythagoars' theorem for finding AQ and QP Then [ OPQ] is calculable and the expression in line 3 *** woold then give the answer. But this seems too laborious, so as I often do, I follow Math Booster's video and read other people's comments.

[@ناصريناصر-س4ب]

In a perpendicular and homogeneous where D(0,0),A(0,6),B(6,6),C(6,0),Q(4,0),P(0,3) the equation of the line BP is y=(1/2)x+6 and the equation of the line AQ is y=(-3/2)x+6, their point of intersection is O(3/2,15/4), so the area of the triangle AOP is (3*3/2)/2=9/4 and the area of the triangle ABP is (3*6)/2=9 and the area of the triangle ADQ is (6*4)/2=12 and from there the red area is 36-(9+12-9/4)=69/4

[@jimlocke9320]

The area of square ABCD is (6)(6) = 36. The area of ΔADQ is (1/2)bh = (1/2)(4)(6) = 12. The base of ΔABO is AB = 6. The height of ΔABO is the difference of the y coordinate of O, which is 15/4, and the y coordinate of line AB, which is 6 = 24/4, so the height is 9/4. The area of ΔABO is (1/2)(6)(9/4) = 27/4. Red area = area of square - area ΔADQ - area ΔABO = 36 - 12 - 27/4 = 69/4.

[@himo3485]

AOB∞QOR 6 : 10 = 3 : 5 3h+5h=6 8h=6 h=3/4 3h=9/4 area of BCQO = (6+2)*6*1/2 - 6*9/4*1/2 = 24 - 27/4 = 96/4 - 27/4 = 69/4

[@MarieAnne.]

Side length of square = 4 + 2 = 6 → AP = PD = 3 Using coordinate geometry with D at origin, we get: A = (0, 6), Q = (4, 0) → AD: y = −(3/2)x + 6 B = (6, 6), P = (0, 3) → BP: y = (1/2)x + 3 The point of intersection of these lines is O (1/2)x + 3 = −(3/2)x + 6 → 2x = 3 → x = 3/2 → y = (1/2)(3/2) + 3 = 15/4 O = (3/2, 15/4) Join points O and C to form △OCQ and △OBC We can use coordinates of O to find altitudes of these two triangles: △OCQ has base CQ = 2 and altitude = 15/4 △OBC has base BC = 6 and altitude = 6−3/2 = 9/2 Area(BCQO) = Area(△OCQ) + Area(△OBC) = (1/2 * 2 * 15/4) + (1/2 * 6 * 9/2) = 15/4 + 54/4 = 69/4

[@五十嵐特許事務所]

Draw a line parallel to DC from point P, and let R be the point of intersection with AQ. AP/AD＝1/2, so PR/DQ＝1/2. DQ＝4, so PR＝2. Also, AB and PR are parallel, so ⊿AOB∽⊿ROP. ∴PO:OB＝PR:AB＝2:6＝1:3. ∴[AOP]:[ABO]＝1:3. ∴[ABO]＝(3/4)×[APB]=(3/4)×9=27/4. Furthermore, [AQD]＝4×6/2＝12. Therefore, [BCQO]=[ABCD]-[ABO]-[AQD]=36-27/4-12=69/4(answer)

[@RobertHering-tq7bn]

Used lines thru O in (x,y) with D(0,0); P(0,3), B(6,6), A(0,6), Q(4,0) and looking for O(Ox,Oy). We have the following for the two lines thru O... (y-3)/(x-0) = (6-3)/(6-0) = 3/6 = 1/2 y = (1/2)x + 3 (y-0)/(x-4) = (0-6)/(4-0) = -6/4 = -(3/2) y = -(3/2)x + 6 This means for Ox we have... (1/2)Ox + 3 = -(3/2)Ox + 6 2Ox = 3 Ox = 3/2 [BCQO] = [BCDP] - [ADQ] + [APO] = (1/2)*6*(6+3) - (1/2)*4*6 + (1/2)*3*(3/2) = 27 - 12 + 9/4 = 15 + 9/4 = 60/4 + 9/4 = 69/4

[@quigonkenny]

As DC = DQ+QC = 4+2 = 6 and ABCD is a square, then all sides of ABCD are of length 6. Set up a coordinate system where the origin is D, the positive vertical (y) axis is DA, and the positive horizontal (x) axis is DC. As AQ passes through (0, 6) and (4, 0), then the slope is (0-6)/(4-0) = -6/4 = -3/2 and the y intercept is 6. Thus the equation of AQ is y = -(3x/2)+6. --- [1] As PB passes through (0, 3) and (6, 6), then the slope is (6-3)/(6-0) = 3/6 = 1/2 and the y intercept is 3. Thus the equation of AQ is y = (x/2)+3. --- [2] Equations [1] and [2] intersect (at point O) where their result is equal. As both are linear equations, there is only one such intersection. -3x/2 + 6 = x/2 + 3 6 - 3 = x/2 + 3x/2 2x = 3 x = 3/2 The area of the shaded quadrilateral OBCQ is equal to the total area of the square ABCD, minus the areas of triangles ∆ADQ and ∆BAP, plus the area of triangle ∆POA, which is the overlap of the two aforementioned triangles. Shaded quadrilateral OBCQ: A = AB(BC) - AD(DQ)/2 - AB(AP)/2 + AP(x)/2 A = 6² - 6(4)/2 - 6(3)/2 + 3(3/2)/2 A = 36 - 12 - 9 + 9/4 A = 15 + 9/4 [ A = 69/4 = 17.25 sq units ]

[@ludmilaivanova1603]

interesting method.

[@Z-eng0]

I didn't use areas like here and literally ignored extending AQ to meet BC like here. I got almost every single sub length in this shape to arrive at the answer of 69/4 though. Aside from extending AQ to meet BC, I extended lots of perpendicular lines from O, One that falls on DC, another on AP, another on BC and a forth on AB, and by similarities and pythagoras theorem got almost all the sub distances formed by those lines (took a lot of effort though. I totally appreciate your use of areas to make those equations, though, which makes it more neat and easy most of the time. Only thing my method did better was the fun arriving at the different distances from each point and from where to where and such, otherwise it's a very long and roundabout method.

[@kidas0808]

S=s1-s-s3; s1=a*a=6*6=36; s2=4*6/2=12; s3=3*6/2=9; S= 36-12-9=15

[@AlbaRosa-ui3xs]

Con questo metodo, hai sottratto 2volte A3 ed è sbagliato

[@gnkgaming6363]

I just calculated the whole area of the square, had 36. I calculated the areas of ABD and ADC, then subtracted them from the area of the whole square but I had 15 as my final answer. Please how is my approach wrong?

[@gandautama4141]

redraw with correct scale got 17.333 approximately...

[@wasimahmad-t6c]

17.683281