Math Olympiad Geometry Challenge | Find the length AD | Important Geometry and Algebra Skills

Math Olympiad Geometry Challenge | Find the length AD | Important Geometry and Algebra Skills
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[@florianbuerzle2703]

A faster solution: Let x = BA = CD and y = AD. Then, by the sine theorem, we have x / sin(105°) = y / sin(30°) and therefore x = y (√6 + √2) / 2. From the cosine theorem we get (5 √2)² = x² + y² - 2xy cos(75°). Substituting x with the result from the sine theorem, we get after simplification, 50 = y² (√6 + √2)² / 4. Solving for y and rationalizing the denominator gives y = 5√3 - 5.

[@nehronghamil4352]

Actually, math boosters method is the quickest and slickest

[@murdock5537]

Amazing!

[@zsboya]

More simply (AB = 5*√2 , at 6:00 -): From vertex B, draw a line perpendicular to side AE and denote the intersection by M. Then BM is both an angular bisector ( cos(45°+30°) = sin(15°) = AM/AB -> AM = AB * √(1 - cos(30°))/√2 = 5 * √(4/4 - 2*√3/4) -> AD = AE = 2*AM = 5 * √(2*(2 - √3)) = 5 * √(3+1 - 2*√3) = 5*(√3-1) unit .

[@jimlocke9320]

At 5:50, Math Booster has found that length AB = 5√2. Drop perpendicular from point A to BC and label the intersection as point F. ΔABF is a 30°-60°-90° special triangle. Length AF = half of length AB, or (5/2)(√2). ΔADF is a well known 15°-75°-90° triangle. Its hypotenuse (AD) is (√2)(√3 - 1) times as long as its longer side (AF). So, length AD = (√2)(√3 - 1)(5/2)(√2) = (2)(5/2)(√(3) - 1) = 5(√(3) - 1), as Math Booster also found.

[@quigonkenny]

As ∠ADC is the external angle for ∆BDA at D, ∠ADC = ∠DAB+∠ABD = 45°+30° = 75°. Draw AE, where E is rhe point on DC where ∠DEA = 75°. As ∠ADE = ∠DEA = 75°, ∆EAD is an isosceles triangle, AE = AD, and ∠EAD = 180°-(75°+75°) = 30°. As ∠EAB = ∠BEA = 75°, ∆ABE is an isosceles triangle and BE = BA. As AE = AD, BA = DC, and ∠EAB = ∠ADC = 75°, by SAS, ∆ABE and ∆DCA are congruent. Therefore BA = BE = AC = DC = 5√2, ∠DCA = ∠ABE = 30°, and ∠CAE = 75°-30° = 45°. By the law of cosines: AD² = AC² + DC² - 2(AC)(DC)cos(30°) AD² = (5√2)² + (5√2)² - 2(5√2)²(√3/2) AD² = 50 + 50 - 50√3 = 50(2-√3) AD = √(50(2-√3)) = 5√(4-2√3) AD = 5√(3+1-2√3) = 5(√3-1) ≈ 3.66

[@EnnioPiovesan]

Draw the height AH relative to the base BC and draw a straight line from D which meets AH at a point M such that HDM=60°. We immediately observe that the angle ADM=15° (by difference between HDA=75° and HDM=60°), ADH is a right triangle with the angles HDA=75° and consequently DAH=15°. Setting AB=a, we can observe: a) the triangle AMD is isosceles since DAM=ADM=15°, therefore DM=AM. b) DMH is half of an equilateral triangle, so DM=2DH, MH = √3 DH. c) ABH is half of an equilateral triangle, so AH=AB/2=a/2, BH=√3a /2. Now AH=AM+MH → AB/2= DM+√3 DH → a/2=2DH+√3 DH = (2+√3)DH, then: DH=a/(2(2+√3)) → DH=a(2-√3)/2. But then BD=BH-DH=√3a /2-a(2-√3)/2=(a/2)(√3-2+√3)=a(√3-1). Consequently: BC = BD+DC = a(√3-1) + a = √3a and therefore BC=2BH, so BH=HC. From what has been said it follows that the triangles AHB and AHC are equivalent since they are both rectangles, have AH in common and BH=CH. This shows that the triangle ABC is isosceles and therefore AB=a=5√2. Returning to the triangle DHA and applying the Pythagorean theorem, we have: AD²=AH²+DH² → AD²=(a/2)²+ (a/2)²(2-√3)² → AD² = (a/2)²(1+4+3-4√3)=( a/2)²(8-4√3) So AD²=(a/2)²2(4-2√3) →AD²=(a/2)² 2(√3-1)² →AD=√2 (a/2) (√3- 1), and finally: AD=√2 5√2/2 (√3-1), so AD=5(√3-1)

[@khaledasfour2374]

Well done for those who are looking for details

[@nehronghamil4352]

Alternate Solution: let a = AB = DC draw line from A to perpendicular to BC intersecting at E. Call it h let b = EC and DE = a - b h = a * sin(30) =.5a (1) from law of sines in triangle AED: h / (sin(75) = ( a -b ) / (sin (15) h = sin(75) / sin(15) * (a-b) let C = sin(75) / sin(15) = (1 + 3^.5) / (3^.5 - 1) = 2 + 3^.5 then h = C * ( a - b ) (2) substituting (1) in (2) : .5a = C * (a - b) b = [ (C - .5) * a ] / C = (1 - 1/ (2 * C)) * a (1 - 1/ (2 * C)) * a = 1 - 1 / [ 4 + 2 * 3^.5 ] * a = ( 3^.5 / 2 ) * a (3) Pythagorean Theorem for triangle AEC: h^2 + b^2 = 50 (4) Substitute (1) & (3) in (4): [ .5^2 + (3^.5 / 2) ^2 ] * a^2 = (1) * a^2 = 50 a = 50^.5 h = .5 * a = 2.5 * 2^.5 AD = h / sin(75) = 2.5 * 2^.5 * [ (2 * 2^.5) / (1 + 3^.5)] = 10 / (1 + 3^.5) = 5 * (3^.5 - 1) ~3.66

[@amanueltesfu-hu2eu]

Superb. Great job

[@RahulKumar-id5cq]

Draw a perpendicular AE to BC Now AE=ABsin30 AD=AE/cos15 =2ABsin 15 DE=ADsin15=2ABsin15.sin15 Now CE=CD-DE =AB-2AB sin15.sin15 Now apply Pythagoras theorem in triangle AEC. and we get AB.

[@forcelifeforce]

You need proper units. For instance, AE = ABsin(30 *degrees*). Do not use a decimal point for multiplication. Use parentheses, brackets, braces, or an asterisk. You stated at the end " ...and we get AB," but the goal is to calculate the length of AD.

[@dafashi666]

你的解法简洁

[@RövşənHəbibov-b6g]

Gox uzun gedişlər edir.