Thank you for another nice problem. We are given a diagram where AP appears to be greater than X with a tangent sloping down from left to right. Usually nothing is assumed from the diagram unless specifically stated to begin with. OT = 2 and TQ = X where O is the midpoint of AB and PQ touches the semicircle at T and QB is the other tangent from Q. PA = 5 - X = PT , tangents from P. Triangle POQ has a right angle at O, because AP is parallel to BQ, so angles at P and Q are supplementary and their halved angles add to 90 degrees By Pythagoras' theorem PO.PO + QO.QO = 5.5 PA.PA +OA.OA = PO.PO and QB.QB + OB.OB = OQ.OQ and these can give a quadratic equation to solve for X. OQ.OQ =X.X + 2.2 OP.OP = (5 - X)(5 - X) + 2.2 so X.X + 4 + (25 +X.X-10X) + 4 = 5.5 collecting terms , and taking 25 away both sides 2.X.X - 10X + 8 =0 X.X -5X + 4 = 0 X= (1/2)( 5 +/ - sqrt(25 - 16)) X = 4 or X = 1 Thinking about this; { either} PA = 4 and X = 1 { or X =4 and PA = 1 in which case the diagram is a very poor illustration.} X = 1 { or 4 }
Простая задачка. Из равенства касательных видим, что РА=5-х. Подымаем АВ выше - ещё минус х, т. е. 5-2х. Что мы видим теперь? - Правильно! Прямоугольный треугольник с гипотенузой 5 и большим катетом 4. Сразу становится понятно, что это за треугольник. Таким образом 5-2х=3, откуда -2х=-2⇔х=1.
Alternatively triangle POQ is right angled in O and, called BQ "x", by 2nd Euclid's theorem x(5-x)=2^2. That gives x=1 or x=4. Supposed (in the figure) BQ x=BQ=1 and AP=4
Drop a perpendicular QT onto AP. Then PTQ ist a 3-4-5 right triangle with PT = 3. Then reflect ABQP at AB, so P'Q'QP is a tangential quadrilateral with incircle and area = r∙s = 2(4x + 16)/2 = 4x + 16. P'Q'QP ist also a trapezoid with area = (4x +6)/2 ∙ 4 = 8x + 12. So solving 8x + 12 = 4x +16 for x results in x = 1.
Trazamos los segmentos PO, OM y OQ y obtendremos 3 triángulos rectángulos: PMO, OMQ y POQ MQ=QB=X PM = 5-X En el triángulo PMO (5-X)² + 2² = PO² En el triángulo OMQ 2² + X² = OQ² En el triángulo POQ PO² + OQ² = PQ² (5-X)² + 2² + 2² + X² = 25 25-10X+X²+8+X² = 25 2X²-10X+8=0 X² - 5X + 4 = 0 X1=4, se rechaza porque no puede tener un valor superior al radio. X2 = 1 X = 1 Saludos
As someone else has already shown here, the problem has 2 valid solutions, x=1 and x=4. Although this is not apparent from the sketch presented, the trapezoid ABQP can have either the left or the right side equal to 4 (and the other equal to 1) if all possible configurations are considered. The condition imposed by the presenter for the segment PN (implying that 5-2x must be positive) to eliminate the x=4 solution is unwarranted, since this segment does not represent an absolute length value (which must indeed be positive), but a relationship between two segments whose sum is positive and constant, and in this context a negative value means that one of the segments is smaller than the other one by complementarity (not negative in absolute value). So, the solution x=4 is valid in the configuration where the left side of the trapezoid is smaller than the right side, which is the mirror image of the original trapezoid.
You went so far and made unnecessary calculatioss. Write triangle with one side equals to 4 and hyperteneous equal to 5 the misSing side length is 3 which is also equal to 5 − 2X. From there you could get the X value which is 1. So easy
In the figure, it looks like PA＞QB, i.e. 5-X＞X, but there is no such restriction in the problem. Therefore, the problem is valid even when X=4. If we only consider the case where X=1, then the answer is not complete.
PQN is a 3,4,5, right triangle then PN = 3
BQ = MQ = AN = x (two tangents theorem)
AP = PM = 3 + x
being PQ = 5 it follows
PQ = x + 3 + x = 5
x = 1
Let G be the point of contact with the semicircle PG=PA=3+x ,QB=QG=x from which PQ= 3+x+x=5, so x=1
Exactly
6:09 PNQ triángulo notable de 37°,53° y 90°.
Lados : 3, 4 y 5
Entonces : 5-2x = 3
5-3 = 2x
2 = 2x
1 = x
Life is easy 👍
I solved for x in the right triangle with sides of (5 - 2x), 4, and 5. With this, x = 1.
X/2=2/(5-x)
5x-x²=4
X²-5x+4=0
(X-1)(X-4)=0
X=1(x
Similarity of right triangles:
x/r = r/(5-x)
x(5-x) = r² = 2² = 4
x² - 5x + 4 = 0 --> x=1 cm (Solved √)
No need of a second grade equation.
Triangle PNQ is a well known 3-4-5 one so that PN = 5 - 2X = 3.
Hence, *X = 1*
半円の中心をOとする。PQと円の接点をR、P、Q、RからOに直線を引く。∠APO₌αとすると∠BQR₌2π₋α
△BOQと△ROQは合同より∠BQO₌∠RQO ∠BQO₌(π₋α)/2₌π/2₋α/2よって∠BOQ₌αより△APO∽△BOQ
AP₌YとおくとY：2₌2：X Y₌4/X PR₌Y₌4/X、QR₌XよりX₊4/X₌5 X^2₋5X₊4₌(X₋4)(X₋1)₌0
X＜2よりX₌1
Thank you for another nice problem.
We are given a diagram where AP appears to be greater than X with a tangent sloping down from left to right.
Usually nothing is assumed from the diagram unless specifically stated to begin with.
OT = 2 and TQ = X where O is the midpoint of AB and PQ touches the semicircle at T and QB is the other tangent from Q.
PA = 5 - X = PT , tangents from P.
Triangle POQ has a right angle at O, because AP is parallel to BQ, so angles at P and Q are supplementary and their halved angles add to 90 degrees
By Pythagoras' theorem PO.PO + QO.QO = 5.5 PA.PA +OA.OA = PO.PO and QB.QB + OB.OB = OQ.OQ and these can give a quadratic equation to solve for X.
OQ.OQ =X.X + 2.2 OP.OP = (5 - X)(5 - X) + 2.2 so X.X + 4 + (25 +X.X-10X) + 4 = 5.5
collecting terms , and taking 25 away both sides 2.X.X - 10X + 8 =0 X.X -5X + 4 = 0 X= (1/2)( 5 +/ - sqrt(25 - 16)) X = 4 or X = 1
Thinking about this; { either} PA = 4 and X = 1 { or X =4 and PA = 1 in which case the diagram is a very poor illustration.}
X = 1 { or 4 }
Простая задачка. Из равенства касательных видим, что РА=5-х. Подымаем АВ выше - ещё минус х, т. е. 5-2х. Что мы видим теперь? - Правильно! Прямоугольный треугольник с гипотенузой 5 и большим катетом 4. Сразу становится понятно, что это за треугольник. Таким образом 5-2х=3, откуда -2х=-2⇔х=1.
Great👍
This is easiest way to solve it. I did the sane
@@nexen1041 египтяне жили давно, но здорово облегчили потомкам жизнь. )
Alternatively triangle POQ is right angled in O and, called BQ "x", by 2nd Euclid's theorem x(5-x)=2^2. That gives x=1 or x=4. Supposed (in the figure) BQ x=BQ=1 and AP=4
Drop a perpendicular QT onto AP. Then PTQ ist a 3-4-5 right triangle with PT = 3. Then reflect ABQP at AB, so P'Q'QP is a tangential quadrilateral with incircle and area = r∙s = 2(4x + 16)/2 = 4x + 16. P'Q'QP ist also a trapezoid with area = (4x +6)/2 ∙ 4 = 8x + 12. So solving 8x + 12 = 4x +16 for x results in x = 1.
Join QO.
∆QBO is Similar to ∆OMP
x/2 = 2/(5-x)
x²-5x+4=0
x = 1, discarding its other root.
√[5^2-4^2]=3
3+x=5-x 2x=2 x=1
AB=2*2=4 ; PQ=5=(5-X)+X---> Proyección vertical de PQ=3---> PA=5-X=3+X---> X=1.
Gracias y un saludo cordial
It can be done in two step.PA and pM are equal and QB and QM are equal. So x=1
Trazamos los segmentos PO, OM y OQ y obtendremos 3 triángulos rectángulos:
PMO, OMQ y POQ
MQ=QB=X
PM = 5-X
En el triángulo PMO
(5-X)² + 2² = PO²
En el triángulo OMQ
2² + X² = OQ²
En el triángulo POQ
PO² + OQ² = PQ²
(5-X)² + 2² + 2² + X² = 25
25-10X+X²+8+X² = 25
2X²-10X+8=0
X² - 5X + 4 = 0
X1=4, se rechaza porque no puede tener un valor superior al radio.
X2 = 1
X = 1
Saludos
The mean "radius" helf of the distance of the longest line that accross the circle
Here, in this example, the radius is 2,
So that, AO=1 OB=1
longest line across is diameter which is two times radius Radius = OA = OB = 2 (given) AB = 4
As someone else has already shown here, the problem has 2 valid solutions, x=1 and x=4. Although this is not apparent from the sketch presented, the trapezoid ABQP can have either the left or the right side equal to 4 (and the other equal to 1) if all possible configurations are considered. The condition imposed by the presenter for the segment PN (implying that 5-2x must be positive) to eliminate the x=4 solution is unwarranted, since this segment does not represent an absolute length value (which must indeed be positive), but a relationship between two segments whose sum is positive and constant, and in this context a negative value means that one of the segments is smaller than the other one by complementarity (not negative in absolute value). So, the solution x=4 is valid in the configuration where the left side of the trapezoid is smaller than the right side, which is the mirror image of the original trapezoid.
(5 - 2x)^2 + 4^2 = 5^2
5 - 2x = 3
x = 1
別解法 ∠ＰＯＡ＝∠ＰＯＭ＝ａ、∠ＭＯＱ＝∠ＱＯＢ＝ｂ、２ａ＋２ｂ＝１８０、a+b＝90。ＰＯ二乗＋ＯＱ二乗＝ＰＱ二乗。
２５＝（５－ｘ）二乗＋４＋４＋ｘｘ。０＝（ｘ－４）（Ｘ－１）。ｘ＜2。Ｘ＝１。
5-x-x=3...x=1
You went so far and made unnecessary calculatioss. Write triangle with one side equals to 4 and hyperteneous equal to 5 the misSing side length is 3 which is also equal to 5 − 2X. From there you could get the X value which is 1.
So easy
1
In the figure, it looks like PA＞QB, i.e. 5-X＞X, but there is no such restriction in the problem. Therefore, the problem is valid even when X=4. If we only consider the case where X=1, then the answer is not complete.