Friday, September 9, 2016

Geometry Problem 1257: Triangle, Altitude, Perpendicular, 90 Degree, Measurement

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1257.

1. Join FD, quad. ACDF is cyclic. Angle
GFD = C,
Also quad. BFEC is cyclic.Hence Angle AFE=Angle JFG= C.
Angle FGD = Angle FGJ = 90
By ASA Tr.FGD is congruent to Tr. FGJ, hence JG=GD
Quad GHDB is cyclic Angle GDH =90-A
And Angle GDB = 90-B, Angle HDB=90-A +90-B=C. We get Angle AGL=C
Since Angle AGL = Angle AFE, JP is parallel to GL, in Tr. JPD, G is midpoint of JD and GL is parallel to JP, hence JP=2×GL
JK + PK =2×GL
JK =2×10-9=11

2. https://goo.gl/photos/6syuJ3KQA1nGxwFk7

Let N is the orthocenter of triangle ABC
We have cyclic quadrilaterals AENF, ECDN, ACDF, BGHD and AEDB ( see sketch)
Let u= ∠ (EBC)= ∠ (DAC)= ∠ (HGD)= ∠ (FEN)
Let v= ∠ (BAD)= ∠ (BCF) and z=∠ (ABE)= ∠ (ACF)
We have ∠ (AFE)= ∠ (JFG)= 90-u => ∠ (FJG)= u => JE//GL
In right triangle ADC we have u+v+z= 90
In right triangle ADG we have ∠ (FDG)+v+z= 90 => ∠ (FDG)= u
Triangle FJD is isosceles => DG/DJ=DL/DP= ½
So JP= 2. GL= 20 => KJ= JP- KP= 11

3. Problem 1257
Let O is orthocenter the triangle ABC, the quadrilateral AFOE,EODC and AFDC are cyclic so
<FEO=<FAO=<FAD=<FCD=<OCD=<OED and triangles KHE ,DHE are equals(<EHK=<EHD=90)so
KH=DH.If EF intersects the circle passing through the points A,E,Dand B(<AEB=90=<ADB) in J’ then arc J’B=arcBD(J’B=BD , <KEH=<DEH=J’EB=<BED).But the diameter of the circle (J’,A,E,D,B) is AB so J’D is perpendicular bisector in AB.Then the points J,J’ coincide so JG=GD.Therefore GL//JP and PJ=2GL=2.10=20.But JK=JP-KP=20-9=11.
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

4. We have a number of cyclic quads and accordingly < B = < GHK = < AEF = < EKD

Hence GL // JP

But also < B = < DEC = < FEA hence Tr.s KHE and DHE are isoceles and KH = HD.

Hence from the midpoint theorem HL = 9/2 and so GH = 11/2 and therefore JK = 11

Sumith Peiris
Moratuwa
Sri Lanka

1. Correction

Tr.s KHE and DHE are congruent not isoceles

5. A shorter cut would have been to realize that if O is the orthocentre of Tr. ABC then O is the incentre of Tr. DEF yielding both tr.s JFD and DEK isoceles.

Either result can then be used along with the mid point theorem to get the value of JK

Sumith Peiris
Moratuwa
Sri Lanka