Math Olympiad | A Very Nice Geometry Problem | 2 Different Methods

[@ناصريناصر-س4ب]

Let

[@ناصريناصر-س4ب]

Let H be the perpendicular projection of the point P on AC, from which HA=HC=8, HB=8-6=2, HA=10-2=8, and we have HA*HB=PH², from which PH=4, and according to the Pythagorean theorem x=√(8²+4²)=4√5

[@TheAlavini]

Your didactic aproach is very good. It clearly show us how to sove a problem usign different points of view. Thank you. Congratulation

[@kateknowles8055]

Thank you , Math Booster, for your help in improving our solving. This is how I thought this problem through: PB is perpendicular with AP and angle PAB equals angle PCB. Angle PCB = arctan PB/10 = theta (say) AQ = QC where PQ is perpendicular with AC so AQ =8 I see similar triangles : APQ and ABP with X/8 = 10/X and this is all that will be needed. X.X = 80 , cross multiplying X = sqrt (80)

[@oscarcastaneda5310]

Brilliant : )

[@kateknowles8055]

@@oscarcastaneda5310 Thank you.

[@Amtar_ul_Jamal]

Tangent Secent theorom: PC×PC = BC×AC PC square = 6×16 x square = 4^6 ~ 9.79

[@זאבגלברד]

Take a circle centered at point O which is the middle between A and B , radius 5 . At a distance of 8 from A , at line x=3 , intersect it with the circle to get point P

[@santiagoarosam430]

M es la proyección ortogonal de P sobre AB---> AC=10+6=16=AM+MC=8+8---> AB=AM+MB=8+2---> Potencia de M respecto a la circunferencia =MP²=2*8=16---> MP=4---> X=√(4²+8²)≠4√5. Gracias y un saludo cordial.

[@holyshit922]

APB is right triangle From definition of cosine of acute angle in triangle APB Let angle PAB = θ cos(θ) = x/10 From cosine law in triangle APC x^2 = 16^2+x^2-2*16*x*cos(θ) x^2 = 16^2+x^2-2*16*x*x/10 16/5x^2=256 x^2 = 16*5 x = 4sqrt(5)

[@Antony_V]

This problem is solvable in the mind: let K be the point perpendicular from P to AB, ABP is a right triangle and ACP isosceles. Since AK=8 and AB=10 by the 1st Euclid's theorem AP^2=AK*AB so x^2=80, x=4√5

[@sakurayayoi-p2r]

別解法　二等辺三角形で底辺に頂角から垂線を落とすと二等分する。したがってＡＭ＝ＭＣ＝8、一方、角ＡＰＢは直角。８、ｈの直角三角形とｈ、（8-6）の直角三角形は相似のため、hh＝16、ｘｘ＝６４＋１６＝８０。Ｘ＝４√5

[@marioalb9726]

R = 10/2 = 5 cm Pytagorean theorem, twice: h² = R²- (½16-R) = 5²- 3²= 4² h² = x² - (½16)² Equalling: x² - 8² = 4² x = 4√5 cm ( Solved √ )

[@johnbrennan3372]

join p to b. In the triangle pmb, pm= sqroot (x^2-64). pb=sqroot(100-x^2) since angle apb=90degrees and mb=2. Then using pythag. We get x^2-64+4=100-x^ 2. So x=4root5

[@marcelomatos2584]

This is easy! Metric relations of the right triangle APB. AP²=AP’•AB ⇒ x²=8•10 =4²•5 ⇒ x=4√5

[@yakupbuyankara5903]

X=4×(5^(1/2))

[@eylultahmaz5029]

Angle similarity✌🏻 solved this

[@RealQinnMalloryu4]

(10)^2=100 180°ABC/100=1.80ABC 1.2^40 1.2^20 1.2^10 1.2^5 1.2^1 2^1(ABC ➖ 2ABC+1).

[@aromaclinic4112]

AM=8 OM=3 MB=2 OP=5 MP=4 x=AP=2√5

[@cyruschang1904]

(10^2 - x^2) - (x^2 - 8^2) = 2^2 x^2 = (10^2 + 8^2 - 2^2)/2 = (100 + 64 - 4)/2 = 80 x = √80 = 4√5

[@marioalb9726]

R = 10/2 = 5 cm Similarity of triangles: x/5 = (10+6)/x x² = 16*5 = 80 x = 4√5 cm ( Solved √ )

[@pedrincalentino6063]

Es la forma más sencilla de resolverlo😊

[@oscarcastaneda5310]

I solved x/10 = 8/x giving x = 4√5.

[@AmirgabYT2185]

x=4√5≈8,98

[@giuseppemalaguti435]

PAB=α ..cosα=x/10...x^2=(10sinα)^2+36-2*6*10sinαcos(90+α)..x^2=220(sinα)^2+36=256-(11/5)x^2...x^2=80

[@Sllow118168]

It will be easier if you use similar triangles.

[@harikatragadda]

PM²= AM*MB=8*2=16 X² = AM²+PM² =8²+16=80 X = 4√5

[@peterbuchli9332]

Thaleskreis ergibt rechten Winkel.... nachher Trigonometrie

[@ina-j2p]

半円の半径が5 ACの中点は半円の中心よりC寄りに3の位置にある. Pの高さは4 ∴x=√｛4^2+(2+6)^2｝ =√(16+64) =√80 =4√5

[@marcgriselhubert3915]

Third method: We use an orthonormal center A and first axis (AB). The equation of the semi circle is (x - 5)^2 + y^2 = 0 or x^2 + y^2 - 10.x = 0. The triangle APC is isosceles, so the abscissa of P is (0 + (10+6))/2 = 8, its ordinate is y with 8^2 + y^2 -10.8 = 0 so y^2 = 80 - 64 = 16 and y = 4 as y is positive, so we have P(8; 4) and VectorOP(8; 4). Then X^2 = OP^2 = 8^2 + 4^2 = 64 + 16 = 80 Finally X = sqrt(80) as X is positive, so X = 4.sqrt(5).