Can you find the length X? | (How to think outside the Box) |

Learn how to find the length X. Important Geometry and Algebra skills are also explained: Pythagorean theorem; isosceles Triangles; congruent Triangles. Step-by-step tutorial by PreMath.com
Today I will teach you tips and tricks to solve the given olympiad math question in a simple and easy way. Learn how to prepare for Math Olympiad fast!
Step-by-step tutorial by PreMath.com
• Can you find the lengt...
Need help with solving this Math Olympiad Question? You're in the right place!
I have over 20 years of experience teaching Mathematics at American schools, colleges, and universities. Learn more about me at
/ premath
Can you find the length X? | (How to think outside the Box) | #math #maths | #geometry
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
#FindX #Length #Triangles #CongruentTriangles #GeometryMath #PythagoreanTheorem
#MathOlympiad #IntersectingChordsTheorem #RightTriangle #RightTriangles
#PreMath #PreMath.com #MathOlympics #HowToThinkOutsideTheBox #ThinkOutsideTheBox #HowToThinkOutsideTheBox? #FillInTheBoxes #GeometryMath #Geometry #RightTriangles
#OlympiadMathematicalQuestion #HowToSolveOlympiadQuestion #MathOlympiadQuestion #MathOlympiadQuestions #OlympiadQuestion #Olympiad #AlgebraReview #Algebra #Mathematics #Math #Maths #MathOlympiad #HarvardAdmissionQuestion
#MathOlympiadPreparation #LearntipstosolveOlympiadMathQuestionfast #OlympiadMathematicsCompetition #MathOlympics #CollegeEntranceExam
#blackpenredpen #MathOlympiadTraining #Olympiad Question #GeometrySkills #GeometryFormulas #Angles #Height #ComplementaryAngles
#MathematicalOlympiad #OlympiadMathematics #CompetitiveExams #CompetitiveExam
How to solve Olympiad Mathematical Question
How to prepare for Math Olympiad
How to Solve Olympiad Question
How to Solve international math olympiad questions
international math olympiad questions and solutions
international math olympiad questions and answers
olympiad mathematics competition
blackpenredpen
math olympics
olympiad exam
olympiad exam sample papers
math olympiad sample questions
math olympiada
British Math Olympiad
olympics math
olympics mathematics
olympics math activities
olympics math competition
Math Olympiad Training
How to win the International Math Olympiad | Po-Shen Loh and Lex Fridman
Po-Shen Loh and Lex Fridman
Number Theory
There is a ridiculously easy way to solve this Olympiad qualifier problem
This U.S. Olympiad Coach Has a Unique Approach to Math
The Map of Mathematics
mathcounts
math at work
Pre Math
Olympiad Mathematics
Two Methods to Solve System of Exponential of Equations
Olympiad Question
Find Area of the Shaded Triangle in a Rectangle
Geometry
Geometry math
Geometry skills
Right triangles
imo
Competitive Exams
Competitive Exam
Calculate the length AB
Pythagorean Theorem
Right triangles
Intersecting Chords Theorem
coolmath
my maths
mathpapa
mymaths
cymath
sumdog
multiplication
ixl math
deltamath
reflex math
math genie
math way
math for fun
Subscribe Now as the ultimate shots of Math doses are on their way to fill your minds with the knowledge and wisdom once again.

[@abeonthehill166]

Another well explained and step by step detailed example of how to solve ! Thanks for sharing Professor , you are my favourite internet Maths Tutor !

[@PreMath]

Wow, thanks!🙏❤️ Glad to hear that! You are very welcome!

[@PreMath]

Wow thanks!🙏❤️

[@sandhyaprabhu6996]

Same here. I feel him to be a great math teacher who makes us think out of box

[@DB-lg5sq]

شكرا لكم على المجهودات يمكن استعمال BCE=a , DCA =45-m , CE=b, CD=c sin45 /x = sin (m+45) /b =sin(90-m) /c sin(45-m)/28 = sin45/b sinm/21=sin45/c ........ 1/x =cos2m /28 1/x=sin2m /21 x^2=28^2+21^2

[@egillandersson1780]

Great one !!! Thank you

[@jimlocke9320]

Drop a perpendicular from C to AB and label the intersection as point F. Right isosceles ΔABC is divided into 2 smaller congruent right isosceles triangles, ΔACF and ΔBCF. Let DF have length y. Then, AF = CF = BF = y + 21 and EF = y - 7. Let

[@PreMath]

Excellent! Thanks for sharing ❤️

[@RK-tf8pq]

I too did it on the same principle, as that seems to be the most obvious way to proceed.

[@dharmendrakhetia3640]

Hey how AF is equal to CF ?

[@giuseppemalaguti435]

CB=a...49+x=√2a. BCE=α..teorema dei seni 28/sinα=a/sin(135-α)..21/sin(45-α)=a/sin(α+90)..divido ,elimino a,risulta tg2α=4/3..calcolo a=28sin(135-α)/sinα=42√2..x=(42√2)√2-49=84-49=35

[@PreMath]

Excellent! Thanks for sharing ❤️

[@dalibormakovnik5717]

Very convoluted but also exciting.😊

[@yalchingedikgedik8007]

That’s very useful and nice Thanks Sir for your efforts Very good explain With my respects ❤❤❤❤❤

[@PreMath]

Glad to hear that! You are very welcome! Thanks for the feedback ❤️

[@MarieAnne.]

Here's my inside-the-box solution: Since △ABC is a right isosceles triangle (∠A = ∠B = 45°, ∠C° = 90) then AC = BC = AB/√2 = (x+49)/√2 Let ∠ACD = α, ∠BCE = β Exterior angle theorem: The measure of an exterior angle is equal to the sum of the two non-adjacent interior angles Therefore: ∠CDE = ∠ACD + ∠DAC = α + 45° ∠CED = ∠BCE + ∠EBC = β + 45° Comparing △ ACE to △ BDC we get ∠CAE = ∠DBC = 45° ∠ACE = ∠BDC = α + 45° ∠AEC = ∠BCD = β + 45° Therefore △ ACE ~ △ BDC and so AE/BC = AC/BD (x+21) / ((x+49)/√2) = ((x+49)/√2) / (x+28) (x+21) (x+28) = (x+49)/√2 * (x+49)/√2 2 (x² + 49x + 21*28) = (x+49)² 2x² + 98x + 2(3*7)(4*7) = x² + 98x + 49² 2x² + 24*49 = x² + 49² 2x² − x² = 49² − 24*49 x² = 49 (49−24) = 49 * 25 x = √(49 * 25) = 7 * 5 *x = 35*

[@jamestalbott4499]

Thank you!

[@PreMath]

You're welcome! Thanks for the feedback ❤️

[@rabotaakk-nw9nm]

4:00-5:30 DA/AE'=21/28=3/4 => => ΔDAE'(3/4/5)×7 => DE'=5×7=35

[@harikatragadda]

Rotate the triangle by 90° anti-clockwise about C. This forms a big Right triangle containing ∆CDE' Congruent to ∆CDE, with X being the hypotenuse of the corner Right triangle with sides 28 and 21. This is a Pythagorean triple, hence X =35

[@PreMath]

Thanks for sharing ❤️

[@unknownidentity2846]

Let's find x: . .. ... .... ..... ABC is an isosceles right triangle. Therefore we can conclude: ∠BAC = ∠ABC = (180° − ∠ACB)/2 = (180° − 90°)/2 = 90°/2 = 45° Let M be the midpoint of AB. Since ABC is an isosceles triangle, the triangles ACM and BCM must be congruent right triangles and we obtain: ∠ACM = 180° − ∠AMC − ∠CAM = 180° − 90° − 45° = 45° = ∠CAM ⇒ AM = CM ∠BCM = 180° − ∠BMC − ∠CBM = 180° − 90° − 45° = 45° = ∠CBM ⇒ BM = CM CM = AM = BM = AB/2 The triangles CDM and CEM are also right triangles, so we can conclude: 1 = tan(45°) = tan(∠DCE) = tan(∠DCM + ∠ECM) = [tan(∠DCM) + tan(∠ECM)]/[1 − tan(∠DCM)*tan(∠ECM)] = [(DM/CM) + (EM/CM)]/[1 − (DM/CM)*(EM/CM)] = CM*(DM + EM)/(CM² − DM*EM) CM = AB/2 = (AD + DE + EB)/2 = (21 + x + 28)/2 = (x + 49)/2 DM + EM = DE = x Now we have to find expressions for DM and EM: AM = AD + DM ⇒ DM = AM − AD = (x + 49)/2 − 21 = (x + 49)/2 − 42/2 = (x + 49 − 42)/2 = (x + 7)/2 BM = BE + EM ⇒ EM = BM − BE = (x + 49)/2 − 28 = (x + 49)/2 − 56/2 = (x + 49 − 56)/2 = (x − 7)/2 Now we are able to calculate the value of x=DE: 1 = CM*(DM + EM)/(CM² − DM*EM) 1 = [(x + 49)/2]*x/{[(x + 49)/2]² − [(x + 7)/2]*[(x − 7)/2]} 1 = 2*(x + 49)*x/[(x + 49)² − (x + 7)*(x − 7)] 1 = (2*x² + 98*x)/[(x² + 98*x + 2401) − (x² − 49)] 1 = (2*x² + 98*x)/(x² + 98*x + 2401 − x² + 49) 1 = (2*x² + 98*x)/(98*x + 2450) 98*x + 2450 = 2*x² + 98*x 2450 = 2*x² 1225 = x² ⇒ x = √1225 = 35 Best regards from Germany

[@PreMath]

Excellent! Thanks for sharing ❤️

[@marcgriselhubert3915]

Without thinking outside the box: We use an orthonormal center C and first axis (CA). We have C(0; 0) A(c; 0) B(0; c) with c = CA = CB *Let t = angleCAD and u = tan(t). The equation of (CD) is y = u.x and the equation of (AB) is y = c -x We have easily the intersection D: D(c/(1+u); (c.u)/(1+u)) and VectorAD((-c.u)/(1-u); (c.u)/(1-u)), then AD = (sqrt(2).c.u)/(1+u) = 21 (equation a) *Let t' = angleACE = t + 45° and v = tan(t') = (1+u)/(1-u). The equation of (CE) is y = v.x and then we have the intersection E (replace u by v in the coordinates of D): E(c/(1+v); (c.v)/(1+v)) or E((c/2).(1-u); (c/2).(1+u)), then VectorBE((c/2).(1-u); (c/2).(-1+u)), then BE = sqrt(2).(c/2).(1-u) = 28 (u0, or u = (-4+5)/3 = 1/3. Now we replace u by 1/3 in equation (a) or equation (b) and obtain that c = 84/sqrt(2) Finally in triangle ABC: AB = sqrt(2).c = 84, and x = 84 -21 -28 = 35.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@yuvvrajkperson]

There is another method without having to think outside the box. Triangles cae and cdb are similar So ratio of alpha + 45 and beta + 45 is equal to the ratio of 21+x and 28+x Where alpha can be written as 45-beta Beta can be eliminated to solve for x. The calculation becomes longer though

[@PreMath]

Thanks for the feedback ❤️

[@kristiansolheim3000]

I used the cosine rule but i got 34.6 as result. Merge 28^28+ x^x-2×28×cos45 and 21^21+x^x-21xcos45. Then x become almost 35.

[@LuisdeBritoCamacho]

STEP-STEP RESOLUTION PROPOSAL : 01) 21 = 3 * 7 02) 28 = 4 * 7 03) AB = 21 + X + 28 ; AB = 49 + X 04) EB / AD = 28 / 21 ; EB / AD = 4 / 3 and AD / EB = 21 / 28 ; AD / EB = 3 / 4 05) Drawing a Vertical Line from Point C to AB we get Point F 06) EF = Height 07) Dividing X in Segment "a" and Segment "b". X = a + b and 21 + a = 28 + b 08) 21 + a = 28 + b ; a - b = 28 - 21 ; a - b = 7 ; a = b + 7 or b = a - 7. Conclusion : a > b 09) To keep the same Proportion 4 /3 we must do the following : Use the Ratio of Proportion (4 / 3 ) in order to keep the Equation : 21 + a = 28 + b, True. 10) 21 + a = 28 + b 11) 21 + 4a / 3 = 28 + a 12) 4a / 3 - a = 7 ; 4a / 3 - 3a / 3 = 7 ; a / 3 = 21 ; a = 21 13) a = 21 and b = a - 7 ; b = 21 - 7 ; b = 14 14) a = 21 and b = 14 15) X = a + b ; X = 21 + 14 ; X = 35 16) Checking : 21 + (21) = 28 + (14) ; 42 = 42 17) AB = 49 + X ; AB = 49 + 35 ; AB = 84 lin un Therefore, MY ANSWER : The Segment X equal 35 Linear Units

[@PreMath]

Excellent! Thanks for sharing ❤️

[@michaeldoerr5810]

The answer is x = 35. Golly that is probably one of the easiest "think outside the box" problems I have learned. And I would like to ask are there any "think outside the box" problema that DO NOT make use of auxiliary lines??? I just want to know before ever trying to find more "think outside the box" problems on your channel.

[@michaeldoerr5810]

Also I think that this problem would not work if the angle was NOT 45°. I could be wrong.

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@soli9mana-soli4953]

My strategy was drawing the same triangle on side AC, but yes we can save some lines 😂

[@Just0Me359]

How can I think that thinking!!!! I feel that I am stupid!!! Why that!!!!

[@PreMath]

Hey no one is perfect! Keep persevering... No worries. We are all lifelong learners. That's what makes our life exciting and meaningful!😀

[@misterenter-iz7rz]

Out of my brain 😢

[@wackojacko3962]

@ 2:45 I use the same technique too visualize Mad Magazine's last page cover Fold-In. Turn imaginary into reality. 🙂

[@PreMath]

😀 Thanks for the feedback ❤️